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The largest nN such that 3n divides 50! is:
21
22
20
23
Correct answer
22
Why this answer is correct
Calculation:
Given,
Number, n = 50
Prime number, p = 3
The exponent of 3 in 50! is given by,
∴ The maximum value of n is 22.
Let one focus of the hyperbola and and the corresponding directrix be . If e and l respectively and the length of the latus rectum of H, then 9 (e2 + l) is equal ro:
14
15
16
12
Correct answer
16
Why this answer is correct
Concept:
Hyperbola parameters:
Calculation:
Given:
and
Then,
and
Now,
Therefore,
Hence, the correct answer is Option 3.
The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to
360
45
2520
1820
Correct answer
2520
Why this answer is correct
Calculation:
11111 2 2 2 0 0
No. of sequences
Note: The Sequence can start with 0.
Hence, the Correct answer is Option 3.
Let f : be a twice differentiable function such that (sinx cosy)(f(2x+2y) – f(2x – 2y)) = (cosx siny)(f(2x+2y) + f(2x – 2y)), for all x, . If , then the value of is:
2
-3
3
-2
Correct answer
-3
Why this answer is correct
Calculation:
Given,
We are given the equation:
Taking terms together:
Using identities:
Rewriting,
Let,
and
Then,
(constant)
Now,
Putting :
So,
Now,
∴ The correct answer is Option 2.
Let A = such that det(A) = 0 and . If I denotes 2 × 2 identity matrix, then the matrix (1 + A)8 is:
Correct answer
Why this answer is correct
Solution:
Hence, the Correct answer is Option 4.
The term independent of x in the expansion of
210
150
240
120
Correct answer
210
Why this answer is correct
Calculation:
Simplify the terms inside the bracket:
Subtracting:
So the expression reduces to
General term:
For the term independent of x:
Required term:
The correct answer is 210.
If , then the number of solutions of , is equal to:
12
6
8
10
Correct answer
8
Why this answer is correct
Solution:
Number of solutions = 8.
Hence, the correct answer is Option 3.
Let a1, a2, a3…. be in an A.P. such that the n is :
11
10
18
17
Correct answer
11
Why this answer is correct
Concept:
Arithmetic Progression (A.P.):
Calculation:
Sum of first 12 odd terms in A.P. is
Equate to the given sum,
Multiply both sides by 5,
Now, sum of first n terms = 0
Since d ≠ 0 and n ≠ 0 ,
Hence, the correct answer is Option 1.
If the function f(x) = 2x3 – 9ax2 + 12a2x + 1, where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that p<sub>2</sub> = q, then f(3) is equal to:
55
10
23
37
Correct answer
37
Why this answer is correct
Concept:
Finding local maxima and minima of a cubic function:
where the function attains local maxima or minima.
Calculation:
Given,
First derivative,
Set to find critical points:
Dividing by 6:
Let the roots be and . Then, by Vieta's formulas,
Given condition,
Substitute :
From sum,
From product,
Squaring the sum and equating to product:
Solving for gives .
Then,
Finally, evaluate :
∴ The correct answer is Option 4.
Let z be a complex number such that . If then the maximum distance of k + ik2 from the circle is:
2
3
Correct answer
Why this answer is correct
Calculation:
Point P = (2, 4), center of circle C = (1, 2)
Equation of a circle:
Maximum distance from point Pto the circle is given by:
Hence, the correct answer is Option 1.
If is nonzero vector such that its projections on the vectors and are equal, then a unit vector along is:
Correct answer
Why this answer is correct
Solution:
Calculate magnitudes:
Set equal projections:
From the first equality:
From the second equality:
Solving
∴ a unit vector along
Hence, the correct answer is Option 3.
Let A be the set of all functions f : Z Z and R be a relation on A such that R = {(f, g) : f (0) = g(1) and f (1) = g(0)}. Then R is:
Symmetric and transitive but not reflective
Symmetric but neither reflective nor transitive
Reflexive but neither symmetric nor transitive
Transitive but neither reflexive nor symmetric
Correct answer
Symmetric but neither reflective nor transitive
Why this answer is correct
Solution:
R= {(f,g): f(0) = g(1) and f(1) = g(0)}
Reflexive: (f, f) € R
⇒ f(0) = f(1) and f(1) = f(0) ⇒ must hold
⇒ But this is not true for all functions
So not reflexive
Symmetric: If (f,g) € R ⇒ (g, f) € R
Now, g(0) = f(1) and g(1) = f(0) ⇒ true
∴ symmetric
Transitive: If (f,g) € R and (g,h) € R
⇒ (f,h) € R
Now (f,g) € R= f(0)= g(1) and f(1)= g(0)
⇒ (g,h) € R ⇒ g(0)= h(1) and g(1) = h(0)
⇒ For (f,h) € R we need f(0) = h(1) and f(1) = h(0)
⇒ Now f(0)= g(1) = h(0) nd f(1) = g(0) = h(1)
Hence transitive
Hence, the correct answer is Option 2
For
7
4
6
-1
Correct answer
7
Why this answer is correct
Concept:
Limits and Expansion of Elementary Functions:
Calculation:
Given,
limx→0 [ x2·sin(αx) + (γ−1)·ex<sup>2</sup> ] / [ sin(2x) − βx ] = 3
Numerator:
x2·sin(αx) = x2·(αx − (αx)3/6 + ... ) = αx3 − α3x5/6 + ...
(γ − 1)·ex<sup>2</sup> = (γ − 1)·(1 + x2 + x4/2! + ...) = (γ − 1) + (γ − 1)x2 + ...
So, numerator ≈ (γ − 1) + (γ − 1)x2 + αx3 + ...
Denominator:
sin(2x) = 2x − (2x)3/6 + ... = 2x − 8x3/6 + ...
So, denominator = 2x − βx − (8x3/6) + ... = (2 − β)x − (4/3)x3 + ...
Now rewrite limit as:
limx→0 [ (γ−1) + (γ−1)x2 + αx3 + ... ] / [ (2−β)x − (4/3)x3 + ... ] = 3
Now, use expansion up to x0 term after dividing:
⇒ limx→0 (γ−1) / (2−β)x = 0 unless (γ−1) = 0
So, γ − 1 = 0 ⇒ γ = 1
Then numerator becomes αx3 + ...
Denominator: (2 − β)x − (4/3)x3 + ...
⇒ lim = αx3 / [ (2 − β)x − (4/3)x3 ]
Now multiply numerator and denominator by x−3:
⇒ α / [ (2 − β)x−2 − (4/3) ]
Now take limit x → 0, x−2 → ∞
For limit to be finite, coefficient of x−2 in denominator must be 0
⇒ (2 − β) = 0
⇒ β = 2
Now denominator = −(4/3),
numerator = α ⇒ α / (−4/3) = 3
⇒ α = −4
Given γ = 1, β = 2, α = −4
β + γ − α = 2 + 1 − (−4) = 7
∴ β + γ − α = 7
If the system of linear equations
has infinitely many solutions, then the value of 22 – 9 is :
49
31
43
37
Correct answer
31
Why this answer is correct
Calculation:
Given,
Also,
Putting in previous equation:
Now,
∴ The correct answer is Option 2.
Let and P1 =1, then the quadratic equation having roots is:
x2 – x + 1 = 0
x2 + x – 1 = 0
x2 – x – 1 = 0
x2 + x + 1 = 0
Correct answer
x2 + x – 1 = 0
Why this answer is correct
Concept:
Sequence and Series – Recurrence Relations:
Calculation:
Given:
Let and .
Since , we have:
Next, use the recurrence at :
Therefore, the characteristic equation is:
We want the quadratic whose roots are and .
If satisfy , then:
Hence, the required equation is:
∴ The correct answer is Option 2.
If S and S' are the foci of the ellipse and P be a point on the ellipse, then min(SP.S'P) + max(SP.S'P) is equal to :
9
27
Correct answer
27
Why this answer is correct
Calculation:
PS + PS' = 2 × 3 √2
⇒ b2 = a2 (1 - e2) ⇒ 9 = 18( 1 - e2)
⇒ e =
Diretrix x =
⇒ PS.PS' =
=
⇒ (PS. PS')max = 18 and (PS .PS')min = 9
Sum = 27
Hence, the correct answer is Option 4.
Let the vertices Q and R of the triangle PQR lie on the line QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is then:
Correct answer
Why this answer is correct
Concept:
Coordinate Geometry - Area of Triangle using Vector Approach:
To find the area of triangle PQR, we use the coordinates of points P, Q, and R.
Points Q and R lie on the line given by the parametric equations: .
Distance QR is given as 5 units.
Point P is fixed at (0, 2, 3).
Area of triangle PQR is given by
We find the point M on the line such that vector is perpendicular to the line.
Using the condition of perpendicularity, we solve for the parameter
Calculate the length PM which is the height of the triangle from P to line QR.
Use QR = 5 and height PM to calculate the area.
Calculation:
Given,
Directions of PM:
Directions of line L:
Since is perpendicular ,
Simplifying,
Coordinates of M:
Calculate length PM:
Calculate area of triangle:
Given relation involving m) and n:
Hence, the correct answer is Option 2.
Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the BCD is equal to :
12
Correct answer
Why this answer is correct
Concept:
Area of Tetrahedron from Areas of Perpendicular Triangles:
The relation for the area of triangle BCD is,
Calculation:
Given,
⇒
∴ The area of triangle BCD is , hence the correct answer is Option 3.
)Let \(a \in \mathbf{R} and A be a matrix of order 3x3 such that det(A) = -4 and A + I = , where I is the identity matrix of order 3 x 3.
If det ((a + 1)adj((a–1)A)) is 2m3n, m, n {0,1,2,…..20}, then m + n is equal to :
14
17
15
16
Correct answer
16
Why this answer is correct
A + I =
⇒A =
Given det(A)= −4.
det(A)= 0⋅(0. 1 − 0. 1)−a. (2. 1−0. a)+1. (2. 1−0. a)
= - a. (2)+1. (2) =−2a + 2
Set equal to −4−4-4 -4
⇒ 2a + 2 = -4 ⟹ −2a = −6 ⟹a = 3:
Also
det((a+1) adj ((a−1)A)) =2m3n
Let’s plug in values:
⇒ det((a−1) A) = (a−1)3.det(A) = (3−1)3.(−4)=23.(−4) = 8.(−4) = −32
⇒ det(adj(a - 1)A)) = (-32)2 = 1024
⇒ (a +1)3 = (3 + 1)3 = 64
⇒ det((a+1)adj((a−1)A)) = 64⋅1024 = 65536
Thus, 65536 = 21630
⇒ Then m = 16,n = 0
⇒ m + n = 16 + 0 = 16
Let the focal chord PQ of the parabola y2 = 4x make an angle of 60° with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point , then is equal to :
15
25
30
40
Correct answer
15
Why this answer is correct
Concept:
Focal Chord of Parabola and Circle Properties:
The focal chord of the parabola passes through points on the parabola making a certain angle with the x-axis.
The circle with one diameter as the focal chord touches the y-axis at a certain point.
The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.
The equation of the circle with diameter PS and its intersection with the y-axis is used to find
Calculation:
Given,
The parabola:
Focal chord PQ makes an angle of with positive x-axis.
Using slope
Coordinates of P:
Circle with diameter PS and focus S:
At x = 0,
Calculate
∴ The correct answer is Option 1.
Let [.] denote the greatest integer function. If \int_{0}^{3} \left[ \frac{1}{e^{x-1}} \right] dx = \alpha - \log_e 2, \text{ then } \alpha^3 \text{ is equal to } ____.
Correct answer
Why this answer is correct
Concept:
Greatest Integer Function and Definite Integral:
Calculation:
f(x) = [e1−x] is a decreasing function
f(0) = [e1] = [2.718] = 2
f(1−ln2) = e1−(1−ln2) = eln2 = 2
⇒ boundary point
f(x) = 2 for x ∈ [0, 1−ln2)
f(1) = [e0] = [1] = 1
f(x) = 1 for x ∈ [1−ln2, 1)
f(x) < 1 for x ≥ 1 ⇒ [f(x)] = 0
Now break the integral accordingly:
∫₀³ [e1−x] dx = ∫₀1−ln2 2 dx + ∫1−ln21 1 dx + ∫₁³ 0 dx
⇒ 2(1 − ln2) + (1 − (1 − ln2)) + 0
⇒ 2 − 2ln2 + ln2 = 2 − ln2
Given: ∫₀³ [e1−x] dx = α − ln2
Comparing both sides:
α − ln2 = 2 − ln2 ⇒ α = 2
Now, α3 = 23 = 8
∴ The value of α3 is 8.
Let ƒ : be a thrice differentiable odd function satisfying ƒ'(x)0, ƒ'(x)=ƒ(x),ƒ(0)=0,ƒ'(0)=3. Then 9ƒ(loge3) is equal to _____.
Correct answer
Why this answer is correct
Concept:
Second-Order ODE and Initial Conditions:
Calculation:
Start from the differential equation:
Multiply both sides by f'(x) and integrate:
Apply initial conditions x=0:
Thus:
Separate variables y=f(x):
Use f(0)=0 to find C:
Hence:
At x=ln3), let y=f(ln3). Then:
Set y=f(ln3):
Solve for y:
Therefore:
Hence, the correct answer is 36.
If the area of the region
Correct answer
Why this answer is correct
Concept:
Area of Region:
Calculation:
Given region:
Area calculation involves three integrals:
Evaluating each integral:
Substitute the values:
Given:
Comparing terms,
Hence, the correct answer is 22.
Three distinct numbers are selected randomly from the set {1,2,3,……,40}. If the probability, that the selected numbers are in an increasing G.P. is gcd(m,n) = 1, then m + n is equal to ____.
Correct answer
Why this answer is correct
Given:
Three distinct numbers are selected randomly from the set {1,2,3,…,40}.
The selected numbers are in increasing Geometric Progression (G.P.).
The probability of this event is m/n where \gcd(m,n) = 1
gcd(m,n) 4= 1.
Calculation:
Total number of ways to select 3 distinct numbers from 40 numbers = (403) = 9880
Let the three numbers in increasing G.P. be: a, ar, ar2, where a, r ∈ N
r>1 (since increasing).
Since all numbers are distinct integers between 1 and 40,
⇒ 1 ≤ a< ar < ar2 ≤ 40
Since r is the common ratio, for the sequence to be an increasing G.P. with integers, r must be a positive integer greater than 1
For r=2:
⇒ a ≤
So possible a=1,2,…,10 → 10 sequences
For r = 3
⇒ a ≤ 40/9 ≈ 4
So Possible a= 1, 2, 3, 4 → 4 sequences
For r = 4
⇒ a ≤ 40/16 = 2.5
Possible a = 1, 2
a= 1,2 → 2 sequences
For r = 5
⇒ a ≤ 40/25 = 1.6
Possible a = 1
a=1 → 1 sequence
For r = 6
⇒ a ≤ 40/36 = 1.11
Possible a= 1 → 1 sequence
For r ≥7 a ≤ < 1, no sequences.
Total number of favorable sequences = 10 + 4 + 2 + 1 + 1 = 18
⇒ Now P =
Here, gcd(94940) = 1 so
⇒ m/n = 9/4940
and m + n = 9 + 4940 = 4949
Hence, the Correct answer is 4949.
The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to _____.
Correct answer
Why this answer is correct
Concept:
Circle Equation and Radius:
Calculation:
Center of circle is .
Radius is distance from center to line :
Equation of circle:
Circle passes through point :
Expanding and simplifying:
Factorizing:
Calculating radii:
Absolute difference between squares of radii:
Hence, the correct answer is 768.
Designate whether each of the following compounds is aromatic or not aromatic.
e, g aromatic and a, b, c, d, f, h not aromatic
b, e, f, g aromatic and a, c, d, h not aromatic
a, b, c, d aromatic and e, f, g, h not aromatic
a, c, d, e, h aromatic and b, f, g not aromatic
Correct answer
a, c, d, e, h aromatic and b, f, g not aromatic
Why this answer is correct
CONCEPT:
Aromaticity and Hückel's Rule
EXPLANATION:
Therefore, the correct answer is: a, c, d, e, h are aromatic and b, f, g are not aromatic.
An optically active alkyl halide C4 H9 Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2 . During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :
But-2-yne
Butan-2-ol
Butan-2-one
Butan-1-al
Correct answer
Butan-2-one
Why this answer is correct
CONCEPT:
Reaction Pathway Involving Alkyl Halide → Alkyne → Ketone
EXPLANATION:
Therefore, the IUPAC name of compound [E] is Butan-2-one.
The property/properties that show irregularity in first four elements of group-17 is/are :
(A) Covalent radius
(B) Electron affinity
(C) Ionic radius
(D) First ionization energy
Choose the correct answer from the options given below:
B and D only
A and C only
B only
A, B, C and D
Correct answer
B only
Why this answer is correct
CONCEPT:
Trends in Group-17 Elements (Halogens)
EXPLANATION:
Therefore, the correct answer is: B only
Which of the following graph correctly represents the plots of KH at 1 bar gases in water versus temperature ?
Correct answer
Why this answer is correct
CONCEPT:
Henry’s Law and Temperature Dependence
EXPLANATION:
Therefore, the correct answer is: Option 4
According to Bohr’s model of hydrogen atom, which of the following statement is incorrect?
Radius of 3rd orbit is nine times larger than that of 1st orbit.
Radius of 8th orbit is four times larger than that of 4th orbit.
Radius of 6th orbit is three time larger than that of 4th orbit.
Radius of 4th orbit is four times larger than that of 2nd orbit.
Correct answer
Radius of 6th orbit is three time larger than that of 4th orbit.
Why this answer is correct
CONCEPT:
Bohr’s Model and Radius of Electron Orbits
r ∝ n²
rₙ = r₁ × n²
EXPLANATION:
r₃ / r₁ = (3)² / (1)² = 9 / 1 = 9
Statement 1 is correct.
r₈ / r₄ = (8)² / (4)² = 64 / 16 = 4
Statement 2 is correct.
r₆ / r₄ = (6)² / (4)² = 36 / 16 = 2.25
Statement 3 says “three times larger”, but the actual ratio is 2.25.
Therefore, Statement 3 is incorrect.
r₄ / r₂ = (4)² / (2)² = 16 / 4 = 4
Statement 4 is correct.
Therefore, the incorrect statement is: Radius of 6th orbit is three times larger than that of 4th orbit (Option 3).
Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
The pressure in the vessel B before opening the stopcock is zero.
Correct answer
The pressure in the vessel B before opening the stopcock is zero.
Why this answer is correct
CONCEPT:
Free Expansion of Gas and Thermodynamic Changes
EXPLANATION:
Hence, the pressure in vessel B before opening the stopcock must be zero.
The correct answer is The pressure in the vessel B before opening the stopcock is zero.
A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are :
1400 mm Hg, A
1400 mm Hg, B
600 mm Hg, B
600 mm Hg, A
Correct answer
600 mm Hg, A
Why this answer is correct
CONCEPT:
Raoult's Law and Vapor Pressure of Solutions
EXPLANATION:
B has the higher vapor pressure, so B is the more volatile component.
Δ, having the lower vapor pressure, is the least volatile component.
Therefore, the vapor pressure of pure B is 600 mm Hg, and the least volatile component of the solution is A.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) H2O(l) Consider the above reaction, what mass of CaCl2 will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3 ?
(Given : Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol–1, respectively)
3.908 g
2.636 g
10.545 g
5.272 g
Correct answer
10.545 g
Why this answer is correct
CONCEPT:
Stoichiometry and Limiting Reactant in Chemical Reactions
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
EXPLANATION:
Molar mass of CaCO3 = 40 + 12 + (16 × 3) = 100 g/mol
Moles of CaCO3 = 1000 g / 100 g/mol = 10 mol
Moles of HCl = Molarity × Volume = 0.76 mol/L × 0.250 L = 0.19 mol
Moles of CaCl2 formed = moles of HCl / 2 = 0.19 / 2 = 0.095 mol
Ca = 40, Cl = 35.5 × 2 = 71, Total = 40 + 71 = 111 g/mol
Mass = moles × molar mass = 0.095 × 111 = 10.545 g
Therefore, the mass of CaCl2 formed is 10.545 g.
If equal volumes of AB2 and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2 at 300 K?
(Given Ksp (at 300 K) for AY2 = 5.2 × 10–7)
3.6 × 10–3 M AB2 , 5.0 × 10–4 M XY
2.0 × 10–4 M AB2 , 0.8 × 10–3 M XY
2.0 × 10–2 M AB2 , 2.0 × 10–2 M XY
1.5 × 10–4 M AB2 , 1.5 × 10–3 M XY
Correct answer
2.0 × 10–2 M AB2 , 2.0 × 10–2 M XY
Why this answer is correct
When equal volumes are mixed molarity reduce to half.
CONCEPT:
Qsp = [A+][Y−]2
EXPLANATION:
Conclusion:
Only option 3 gives Qsp > Ksp, so precipitation of AY2 occurs.
Among SO2 , NF3 , NH3 , XeF2 , ClF3 and SF4 , the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is
sp<sub>3</sub>
dsp2
sp3d2
sp3d
Correct answer
sp3d
Why this answer is correct
CONCEPT:
Hybridization, Molecular Geometry, and Dipole Moment
EXPLANATION:
| Molecule | Hybridization | Dipole Moment | Lone Pairs on Central Atom |
|---|---|---|---|
| SO2 | sp² | Non-zero | 1 |
| NF3 | sp³ | Non-zero | 1 |
| NH3 | sp³ | Non-zero | 1 |
| XeF2 | sp³d | Zero | 3 |
| ClF3 | sp³d | Non-zero | 2 |
| SF4 | sp³d | Non-zero | 1 |
Therefore, the correct answer is: sp³d (Option 4).
Given below are two statements :
Statement (I) : Vanillin
will react with NaOH and also with Tollen’s reagent.
Statement (II) : Vanillin
will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct
Correct answer
Statement I is correct but Statement II is incorrect
Why this answer is correct
CONCEPT:
EXPLANATION:
Phenolic group soluble in NaOH
Benzaldehyde derivative react with Tollen’s reagent.
Vanillin does not give self-aldol reaction due to lack of acidic H for condensation.
Therefore, Statement I is correct but Statement II is incorrect.
Identify the correct statement among the following:
All naturally occurring amino acids except glycine contain one chiral centre.
All naturally occurring amino acids are optically active.
Glutamic acid is the only amino acid that contains a –COOH group at the side chain.
Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.
Correct answer
Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.
Why this answer is correct
CONCEPT:
EXPLANATION:
Therefore, the correct statement is: Amino acid cysteine easily undergoes dimerization due to the presence of free SH group.
The correct order of basic nature on aqueous solution for the bases NH3 , H2 N–NH2 , CH3 CH2 NH2 , (CH3CH2 )2NH and (CH3CH2)3 N is :
NH3<H2N–NH2<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NH
NH3<H2N-NH2,<CH3CH2NH2<(CH3CH2),NH<(CH3CH2)3N
H2N–NH2<NH3<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NH
NH2–NH2,<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH
Correct answer
NH2–NH2,<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH
Why this answer is correct
CONCEPT:
EXPLANATION:
Therefore, the correct order of basicity is:
NH2–NH2,<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH
Given below are two statements :
Statement (I) : The metallic radius of Al is less than that of Ga.
Statement (II) : The ionic radius of Al3+ is less than that of Ga3+.
In the light of the above statements, choose the most appropriate answer from the options given below :
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
Correct answer
Statement I is incorrect but Statement II is correct
Why this answer is correct
ONCEPT:
EXPLANATION:
\implies \text{The metallic radius order of Al & Ga is} \ \mathrm{B} < \mathrm{Ga} < \mathrm{Al} < \mathrm{In} < \mathrm{Tl} \ \underbrace{\mathrm{Ga} < \mathrm{Al}}_{\text{(due to poor shielding of d-subshell electrons)}} \ \implies \text{The ionic radius order of } \mathrm{Al}^{+3} & \mathrm{Ga}^{+3} \text{ is} \ \mathrm{B}^{+3} < \mathrm{Al}^{+3} < \mathrm{Ga}^{+3} < \mathrm{In}^{+3} < \mathrm{Tl}^{+3}
Therefore, the correct answer is: Statement I is incorrect but Statement II is correct.
Given below are two statements :
Statement (I) : In octahedral complexes, when < P high spin complexes are formed. When > P low spin complexes are formed. Statement (II) : In tetrahedral complexes because of < P, low spin complexes are rarely formed. In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are correct
Correct answer
Both Statement I and Statement II are correct
Why this answer is correct
CONCEPT:
Explanation:-
Therefore, both Statement I and Statement II are correct.
Choose the correct tests with respective observations.
(A) CuSO4 (acidified with acetic acid) + K4 [Fe(CN)6 ] Chocolate brown precipitate.
(B) FeCl3 + K4 [Fe(CN)6] Prussian blue precipitate.
(C) ZnCl2 + K4 [Fe(CN)6], neutralised with NH4 OH White or bluish white precipitate.
(D) MgCl2 + K4 [Fe(CN)6 ] Blue precipitate.
(E) BaCl2 + K4 [Fe(CN)6 ], neutralised with NaOH White precipitate.
Choose the correct answer from the options given below :
A, D and E only
B, D and E only
A, B and C only
C, D and E only
Correct answer
A, B and C only
Why this answer is correct
CONCEPT:
EXPLANATION:
Therefore, the correct tests with respective observations are A, B, and C only.
On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2 and 0.567 g of H2 O. The empirical formula mass of compound (X) is _________ g. (Given molar mass in g mol–1 C : 12, H : 1, O : 16)
30
45
60
15
Correct answer
30
Why this answer is correct
CONCEPT:
EXPLANATION:
| Component | Given Mass (g) | Molar Mass (g/mol) | Moles (mol) |
|---|---|---|---|
| C (from CO2) | 1.46 (CO2) | 44 (CO2) | 1.46 / 44 = 0.033 |
| H (from H2O) | 0.567 (H2O) | 18 (H2O) | 0.567 / 18 = 0.0315 |
Ratio of moles: C : H : O = 0.033 : 0.063 : 0.0338
Divide by smallest (0.033):
C : H : O ≈ 1 : 1.91 ≈ 2 : 1 : 1 (approximately)
Empirical formula: CH2O
Empirical formula mass = (12 × 1) + (1 × 2) + (16 × 1) = 30 g/mol
Therefore, the empirical formula mass of compound (X) is 30 g/mol.
Consider the following compound (X)
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C – H bond are :
II, IV
III, II
I, IV
II, I
Correct answer
II, I
Why this answer is correct
CONCEPT:
EXPLANATION:
II most stable carbon radical due to resonance stablise
I least stable carbon radical due to no stabilising factor.
Therefore, the most stable radical is at carbon II and the least stable radical is at carbon I.
Consider the following molecules :
The correct order of rate of hydrolysis is :
r > q > p > s
q > p > r > s
p > r > q > s
p > q > r > s
Correct answer
p > q > r > s
Why this answer is correct
CONCEPT:
EXPLANATION:
Rate of hydrolysis Leaving group ability
Therefore, the correct order of rate of hydrolysis is p > q > r > s
A molecule with the formula AX4Y has all it’s elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is :
Square pyramidal
Octahedral
Pentagonal planar
Trigonal bipyramidal
Correct answer
Square pyramidal
Why this answer is correct
CONCEPT:
EXPLANATION:
Therefore, the shape of the molecule AX4Y (XeOF4) is square pyramidal.
A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (M3+/M2+). It forms a metal complex of the type [M(CN)6 ]4–. The number of electrons present in the eg orbital of the complex is ________.
Correct answer
Why this answer is correct
CONCEPT:
EXPLANATION:
Co has highest standard electrode potential (M+3/M+2) among Mn, Cr, Co, Fe
Complex is [Co(CN)6 ]4– and its splitting is as follows.
Therefore, the number of electrons present in the eg orbital of the complex is 1.
Consider the following electrochemical cell at standard condition.
\(\mathrm{Au(s)}|\mathrm{QH}_2,\mathrm{Q}|\mathrm{NH}4\mathrm{X}(0.01\mathrm{M})||\mathrm{Ag}^+(1\mathrm{M})|\mathrm{Ag(s)} \ E{\text{cell}} = +0.4\mathrm{V} \ \text{The couple } \mathrm{QH}_2/\mathrm{Q} \text{ represents quinhydrone electrode, the half cell reaction is given below} \ \)
\left[ \text{Given: } E_{\mathrm{Ag}^+/\mathrm{Ag}}^0 = +0.8\mathrm{V} \text{ and } \frac{2.303\mathrm{RT}}{\mathrm{F}} = 0.06\mathrm{V} \right] \ \text{The } pK_b \text{ value of the ammonium halide salt } (\mathrm{NH}_4\mathrm{X}) \text{ used here is } ____. \text{(nearest integer)}
Correct answer
Why this answer is correct
CONCEPT:
Electrochemical Cells & Nernst Equation
E = Eo - (0.0591/n) log [Products]/[Reactants]
E = Eo - 0.06 log[H+]2
pH = -log[H+]
EXPLANATION:
\(\mathrm{QH}2 + 2\mathrm{Ag}^+ \rightarrow 2\mathrm{Ag} + \mathrm{Q} + 2\mathrm{H}^+ \ E = E^\circ - \frac{0.06}{2} \log [\mathrm{H}^+]^2 \ E = E^\circ - 0.06 \log [\mathrm{H}^+] \ \mathrm{pH} = -\log [\mathrm{H}^+] = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06} \ = \frac{0.3}{0.06} = 5 \ \mathrm{pH}{\mathrm{NH}_4\mathrm{X}} = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log C \ 5 = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log (10^{-2}) \ pK_b = 6.\)
Therefore, the pKb of NH4X is 6.
0.1 mol of the following given antiviral compound (P) will weigh _______ × 10–1 g
(Given : molar mass in g mol–1 H: 1, C : 12, N : 14, O : 16, F : 19, I : 127)
Correct answer
Why this answer is correct
CONCEPT:
Molar Mass and Mole Concept
Mass = Number of moles × Molar mass
EXPLANATION:
Molar mass = 372 gm
So,545 0.1 mole has = 372 ×10–1 gm
Therefore, the answer is 372.
Consider the following equilibrium, CO(g) + 2H2 (g) CH3 OH(g) 0.1 mol of CO along with a catalyst is present in a 2 dm3 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH3 OH is formed. The Kp is _________ × 10–3 (nearest integer). Given : R = 0.08 dm3 bar K–1 mol–1 Assume only methanol is formed as the product and the system follows ideal gas behaviour.
Correct answer
Why this answer is correct
CONCEPT:
Kp = (PCH3OH) / (PCO × PH22)
EXPLANATION:
\(\mathrm{CO(g)} + 2\mathrm{H}2\mathrm{(g)} \rightleftharpoons \mathrm{CH}3\mathrm{OH(g)} \ \begin{array}{lccc} & \mathrm{CO} & 2\mathrm{H}2 & \mathrm{CH}3\mathrm{OH} \ \text{t = 0} & 0.1 , \text{mol} & a , \text{mol} & - \ \text{t}{\text{eq}} & 0.1 - x & a - 2x & x = 0.04 \ & = 0.06 & = a - 0.08 & = 0.04 \ & & = 0.23 - 0.08 & \ & & = 0.15 , \text{mole} & \end{array} \ V = 2L \ T = 500 , \mathrm{K} \ P{\text{total}} = 5 , \text{bar} \ n{\text{Total}} = 0.06 + (a - 0.08) + 0.04 = a + 0.02 \ P{\text{total}} = n_{\text{total}} \times \frac{RT}{V} \ 5 = (a + 0.02) \times \frac{0.08 \times 500}{2} \ 5 = (a + 0.02) \times 0.08 \times 250 \ 5 = (a + 0.02) \times 20 \ \frac{5}{20} = a + 0.02 \ 0.25 = a + 0.02 \ a = 0.25 - 0.02 = 0.23 , \text{mol}\)
\(\implies a = 0.25 - 0.02 = 0.23 , \text{mol.} \ K_p = \frac{X_{\mathrm{CH}3\mathrm{OH}}}{X{\mathrm{CO}} \times X_{\mathrm{H}2}^2} \times \frac{1}{(P_T)^2} = \frac{0.04}{0.06 \times (0.15)^2} \times \left[ \frac{1/4}{5} \right]^2 \ = \frac{0.04}{0.06 \times 0.0225} \times \left[ \frac{1}{20} \right]^2 = \frac{40000}{6 \times 225} \times \frac{1}{400} \ = \frac{100}{6 \times 225} = \frac{100}{1350} = \frac{10}{135} = \frac{2}{27} \ K_p = \frac{0.04}{0.06 \times (0.15)^2} \times \left( \frac{n{\text{total}}}{P_{\text{total}}} \right)^2 = \frac{0.04}{0.06 \times 0.0225} \times \left( \frac{0.25}{5} \right)^2 \ = \frac{0.04}{0.00135} \times (0.05)^2 = \frac{0.04}{0.00135} \times 0.0025 = \frac{40000}{135} \times \frac{25}{10000} \ = \frac{4 \times 25}{135} = \frac{100}{135} = \frac{20}{27} \approx 0.74 \ K_p = \frac{X_{\mathrm{CH}3\mathrm{OH}}}{X{\mathrm{CO}} \times X_{\mathrm{H}_2}^2} \times \frac{1}{P_T^2} = \frac{0.04 / 0.25}{(0.06 / 0.25) \times (0.15 / 0.25)^2} \times \frac{1}{5^2} \ = \frac{0.16}{0.24 \times (0.6)^2} \times \frac{1}{25} = \frac{0.16}{0.24 \times 0.36} \times \frac{1}{25} = \frac{0.16}{0.0864} \times \frac{1}{25} \ = \frac{1600}{864} \times \frac{1}{25} = \frac{64}{864} = \frac{8}{108} = \frac{2}{27} \approx 0.074 = 74 \times 10^{-3}\)
Rounding to nearest integer, Kp ≈ 74 × 10−3.
Therefore, the value of Kp is approximately 74 × 10−3.
For the reaction A products.
The concentration of A at 10 minutes is ________ × 10–3 mol L–1 (nearest integer). The reaction was started with 2.5 mol L–1 of A.
Correct answer
Why this answer is correct
CONCEPT:
Zero Order Reaction Kinetics
[A]t = -Kt + [A]0
t1/2 ∝ [A]0
Slope = 1 / 2K
EXPLANATION:
Slope = 1 / 2K → K = 1 / (2 × 76.92)
K = 1 / 153.84 ≈ 0.0065 min−1
[A]t = -Kt + [A]0
[A]10 = - (1 / (2 × 76.92)) × 10 + 2.5
[A]10 ≈ -0.065 + 2.5 = 2.435 mol L−1
2.435 mol L−1 = 2435 × 10−3 mol L−1
Therefore, the concentration of A at 10 minutes is 2435 × 10−3 mol L−1.
A light wave is propagating with plane wave fronts of the type x + y + z = constant. The angle made by the direction of wave propagation with the x-axis is :
Correct answer
Why this answer is correct
Calculation:
The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y, and z axis.
Angle made by the light with x, y & z axis is same.
Thus, cosα = cosβ = cosγ (α, β, and γ are the angles made by light with x, y, and z axes respectively)
Also, cos²α + cos²β + cos²γ = 1 [Sum of direction cosines]
⇒ α = cos⁻¹(1/√3)
The equation for real gas is given by , where P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab–2 is equivalent to that of :
Planck's constant
Compressibility
Strain
Energy density
Correct answer
Energy density
Why this answer is correct
Calculation:
(P + a/V²)(V - b) = RT
Therefore, [a] = [P][V²] = M L⁻¹ T⁻² L⁶ = M L⁵ T⁻²
[b] = [V] = L³
[ab⁻²] = M L⁵ T⁻² L⁻⁶ = M L⁻¹ T⁻²
⇒ Dimension of energy density
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be :
20 rad/s
30 rad/s
10 rad/s
0 rad/s
Correct answer
20 rad/s
Why this answer is correct
Calculation:
WF = 20 × 1 = 20 J
Therefore, ΔKE = 20 J = (1/2) I ω²
I = M R² = 10 × 0.1² = 0.1 kg m²
Therefore, 20 = (1/2) × 0.1 × ω²
Implying, ω = 20 rad/sec
A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :
Correct answer
Why this answer is correct
Calculation:
Location of image of A :-
(1/v) - (1/u) = (1/f)
⇒ (1/v) - (1/(v - 30)) = (1/20)
⇒ (1/v) = (1/60)
⇒ v = 60 cm
The magnification m = 2.
Since the size of the object is small with respect to the location, hence
dv = m² du ⇒ dv = 4 × 1 = 4 cm
hi = m h0 ⇒ hi(dy) = 2 × 2 = 4 cm
Therefore, Angle made with principle a×is = -45°
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density .The force experienced by a point charge +q placed at the mid point between two plates will be
Correct answer
Why this answer is correct
Calculation:
Given:
Surface charge density on Plate 1, σ1 = +σ
Surface charge density on Plate 2, σ2 = -2σ
Point charge, q = +q at midpoint between plates
Distance between plates = d (charge at midpoint, so at d/2 from each plate)
Electric field due to Plate 1 (positive charge):
⇒ E1 = σ / (2ε0) directed away from Plate 1
Electric field due to Plate 2 (negative charge):
⇒ E2 = 2σ / (2ε0) = σ / ε0, directed toward Plate 2 (since negative charge)
Since charge q is at midpoint, the net electric field Enet is vector sum:
⇒ Enet = E1 (rightward) + E2 (rightward, since Plate 2's field points toward it)
⇒ Enet = (σ / 2ε0) + (σ / ε0) = (3σ) / (2ε0)
Force on point charge q:
⇒ F = q × Enet
⇒ F = q × (3σ) / (2ε0)
∴ Force on charge q = (3σ q) / (2ε0)
A river is flowing from west to east direction with speed of 9 km h–1. If a boat capable of moving at a maximum speed of 27 km h–1 in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150° to direction of river flow, then the width of the river is :
300 m
112.5 m
75 m
112.5 x
Correct answer
112.5 m
Why this answer is correct
Calculation:
Time taken to cross = 30 seconds = 30 / 3600 hr = 1 / 120 hr
Perpendicular velocity, V⊥ = 27 / 2 km/hr
⇒ Width of river = V⊥ × time
⇒ Width = (27 / 2) × (1 / 120) km
⇒ Width = (27 × 1000) / (2 × 120) m
⇒ Width = 112.5 m
∴ Width of the river is 112.5 m
A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is :
(4d/3, 0, 0)
(d/4, 0, 0)
(3d/4, 0, 0)
(d/3, 0, 0)
Correct answer
(d/4, 0, 0)
Why this answer is correct
Calculation:
Let the position where electric field is zero be at distance x from charge +q at origin.
Then, distance from +9q is (d − x)
Equating magnitudes of electric fields,
⇒ (kq) / x2 = (k × 9q) / (d − x)2
⇒ 1 / x2 = 9 / (d − x)2
⇒ (d − x)2 = 9x2
⇒ d − x = 3x
⇒ x = d / 4
∴ The coordinates of point P are (d / 4, 0, 0)
The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged ?
43.8 kJ
48.7 kJ
87.7 kJ
24.4 kJ
Correct answer
87.7 kJ
Why this answer is correct
Calculation:
Given, V = 4.2 V
Battery capacity = 5800 mAh = 5800 × 3600 × 10−3 C
⇒ Q = 5800 × 3.6 = 20880 C
⇒ Energy = V × Q = 4.2 × 20880 = 87700 J
⇒ Energy = 87.7 kJ
∴ Energy stored in the battery when fully charged is 87.7 kJ
A particle is subjected two simple harmonic motions as
where x is displacement and t is time in seconds. The maximum acceleration of the particle is x × 10–2 ms–2. The value of x is :
175
125
Correct answer
175
Why this answer is correct
Calculation:
Given:
x1 = √7 sin 5t cm
x2 = 2√7 sin(5t + π/3) cm
Angular frequency, ω = 5 rad/s
Using phasor diagram:
Angle between x1 and x2 = 60°
Resultant amplitude A:
A = √[ (√7)2 + (2√7)2 + 2 × √7 × 2√7 × cos 60° ]
⇒ A = √[ 7 + 4 × 7 + 2 × √7 × 2√7 × (1/2) ]
⇒ A = √49 = 7 cm
Maximum acceleration:
amax = A × ω2
⇒ amax = 7 × 25 cm/s2
⇒ amax = 175 × 10-2 m/s2
∴ The value of x is 175
The relationship between the magnetic susceptibility and the magnetic permeability is given by :
(0 is the permeability of free space and r is relative permeability)
Correct answer
Why this answer is correct
Calculation:
We are given: μr = 1 + χ
⇒ χ = μr - 1
Also, μ = μ0 μr
⇒ μr = μ / μ0
Substitute into the equation for χ:
⇒ χ = (μ / μ0) - 1
∴ χ = (μ / μ0) - 1
A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor connected in series, the zener current is found to be 4 times load current. The load current (IL) and load resistance (RL) are :
IL = 20 mA; RL = 250
IL = 10 A; RL = 0.5
IL = 0.02 mA; RL = 250
IL = 10 mA; RL = 500
Correct answer
IL = 10 mA; RL = 500
Why this answer is correct
Calculation:
From the circuit diagram,
⇒ 5i = 20 / 400
⇒ 5i = 1 / 20
⇒ i = 1 / 100 = 0.01 A = 10 mA (Load current)
Given load voltage, VL = 5 V
⇒ Load resistance, RL = VL / i
⇒ RL = 5 / (10 × 10-3) = 500 Ω
∴ Load current i = 10 mA and Load resistance RL = 500 Ω
In an adiabatic process, which of the following statements is true ?
The molar heat capacity is infinite
Work done by the gas equals the increase in internal energy
The molar heat capacity is zero
The internal energy of the gas decreases as the temperature increases
Correct answer
The molar heat capacity is zero
Why this answer is correct
Calculation:
For adiabatic process,
⇒ dQ = 0
⇒ Molar heat capacity = 0
⇒ From first law, dQ = dU + dW
⇒ 0 = dU + dW ⇒ dU = -dW
Also,
⇒ dU = (f / 2) × nR × dT
∴ Only option (3) is correct.
A square Lamina OABC of length 10 cm is pivoted at 'O'. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :
20 N
0 (zero)
10 N
Correct answer
10 N
Why this answer is correct
Concept
Explanation
The square lamina of side is pivoted at point .
Forces of act at different points producing clockwise and anticlockwise torques about .
Calculations
Taking moments about point :
Balancing torques for equilibrium:
Thus, the magnitude of the force required for equilibrium is .
Let B1 be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B2 be the magnitude of magnetic field at an axial distance 'x' from the center. For x
4 : 5
16 : 25
64 : 125
25 : 16
Correct answer
64 : 125
Why this answer is correct
Calculation:
Given:
⇒ B1 = (μ0 i) / (2R)
⇒ B2 = B1 × sin3 θ
⇒ Therefore, B2 / B1 = sin3 θ
⇒ sin θ = 4 / 5
⇒ (sin θ)3 = (4 / 5)3 = 64 / 125
∴ The ratio B2 / B1 = 64 / 125.
Considering Bohr's atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of He+ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li++ ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of He+ ion for its ground state.
(D) the energy of He+ ion in its first excited state is same as that for Li++ ion in its ground state
Choose the correct answer from the options given below :
(B), (D) only
(A), (B) only
(A), (D) only
(A), (C) only
Correct answer
(A), (B) only
Why this answer is correct
Concept Used:
Energy levels in hydrogen-like atoms are proportional to the atomic number divided by the square of the principal quantum number
E = k × Z2 / n2 (where k is a constant)
Excited states correspond to n values greater than 1
Explanation:
From the given data:
⇒ For 1st excited state, n = 2
⇒ For 2nd excited state, n = 3
∴ Only statements A and B are correct.
Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is . Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :
/4
/8
/2
Correct answer
/4
Why this answer is correct
Calculation:
Before cutting
After cutting
For each half:
⇒ Mass = M / 2
⇒ Length = ℓ / 2
Moment of inertia for each half, α' (single part) = (M/2) × (ℓ/2)2 / 12
⇒ α' = (M/2) × (ℓ2 / 4) / 12 = (M × ℓ2) / 96
Total moment of inertia, α' = 2 × (M × ℓ2 / 96) = (M × ℓ2) / 48
⇒ α' = α / 4
∴ Correct option is (2).
A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is : (C is the center of curvature of the spherical surface and R is the radius of curvature)
0.24 m right to the spherical surface
0.4 m left to the spherical surface
0.24 m left to the spherical surface
0.4 m right to the spherical surface
Correct answer
0.4 m left to the spherical surface
Why this answer is correct
Calculation:
⇒ (1.5 / v) - (1 / -0.2) = (1.5 - 1) / 0.4
⇒ (1.5 / v) = 1.25 - 5 = -3.75
⇒ v = 1.5 / (-3.75) = -0.4 m
∴ v = -1.2 m.
Match List–I with List–II.
| List–I | List–II |
|---|---|
| (A) Coefficient of viscosity | (I) [ML0T–3] |
| (B) Intensity of wave | (II) [ML–2T–2] |
| (C) Pressure gradient | (III) [M–1LT2] |
| (D) Compressibility | (IV) [ML–1T–1] |
Choose the correct answer from the options given below :
(A)–(I), (B)–(IV), (C)–(III), (D)–(II)
(A)–(IV), (B)–(I), (C)–(II), (D)–(III)
(A)–(IV), (B)–(II), (C)–(I), (D)–(III)
(A)–(II), (B)–(III), (C)–(IV), (D)–(I)
Correct answer
(A)–(IV), (B)–(I), (C)–(II), (D)–(III)
Why this answer is correct
Explanation:
Coefficient of viscosity (η):
Expressed as force per unit area per velocity gradient.
⇒ Dimensional formula: [M1 L-1 T-1].
Intensity (I):
Power delivered per unit area perpendicular to wave propagation.
SI unit is watt per square meter (W/m2).
⇒ Dimensional formula: [M1 L0 T-3].
Pressure gradient:
Represents the rate of change of pressure with distance.
Its unit is pascal per meter (Pa/m).
⇒ Dimensional formula: [M L-2 T-2].
Compressibility (K):
Defined as the fractional change in volume per unit increase in pressure.
Inverse of bulk modulus (elastic modulus of volume).
⇒ Dimensional formula: [M-1 L1 T2].
A small bob of mass 100 mg and charge +10 C is connected to an insulating string of length 1 m. It is brought near to an infinitely long non conducting sheet of charge density '' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be :
(Given, = 8.85 × 10–12 and acceleration due to gravity, g = 10 m/s2)
0.885 nC/m2
17.7 nC/m2
885 nC/m2
1.77 nC/m2
Correct answer
1.77 nC/m2
Why this answer is correct
Calculation:
The FBD is shown below
From equilibrium condition:
⇒ q × (σ / 2 ε0) = mg
⇒ σ = (2 ε0 mg) / q
⇒ σ = (2 × 8.85 × 10-12 × 100 × 10-6 × 10) / (10 × 10-6)
⇒ σ = 17.7 × 10-10 C/m2
⇒ σ = 1.77 nC/m2
∴ Surface charge density σ = 1.77 nC/m2.
A monochromatic light is incident on a metallic plate having work function ϕ . An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is :
(Given : The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)
Correct answer
Why this answer is correct
Calculation:
Using KEmax:
⇒ KEmax = (hc / λ) - φ
Momentum of electron:
⇒ p = √(2m KEmax) = √[2m ((hc / λ) - φ)]
Diameter of path:
⇒ dA-B = 2r = 2 (p / qB) = 2 [√(2m ((hc / λ) - φ)) / eB]
Thus,
⇒ dA-B = (2 √(2m ((hc / λ) - φ))) / (eB)
⇒ dA-B = √[ (8m ((hc / λ) - φ)) / (eB) ]
A vessel with square cross–section and height of 6 m is vertically partitioned. A small window of 100 cm2 with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5 × 103 kg/m3. What force one needs to apply on the hinged door so that it does not get opened ?
(Acceleration due to gravity = 10 m/s2)
Correct answer
Why this answer is correct
Calculation:
Calculate forces:
⇒ Fℓ = (P0 + ρℓ g h) A
⇒ Fw = (P0 + ρw g h) A
So,
⇒ Fext = Fℓ - Fw
⇒ Fext = [ (P0 + ρℓ g h) - (P0 + ρw g h) ] A
⇒ Fext = (ρℓ - ρw) g h A
⇒ Fext = (1500 - 1000) × 10 × 3 × (100 × 10-4)
⇒ Fext = 150 N
∴ The external force required is 150 N.
A steel wire of length 2 m and Young's modulus 2.0 × 1011 Nm–2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10–3 respectively, then the elastic potential energy density of the wire is ____ × 105 (in SI units)
Correct answer
Why this answer is correct
Calculation:
From Poisson's ratio formula:
⇒ μ = - (Δr/r) / (Δℓ/ℓ) ⇒ (Δℓ/ℓ) = - (Δr/r) / μ
⇒ (Δℓ/ℓ) = - (10-3) / 0.2 = - 5 × 10-3
Since length increases, take positive value:
⇒ εℓ = 5 × 10-3
⇒ u = (1/2) × Y × εℓ2
⇒ u = (1/2) × 2 × 1011 × (5 × 10-3)2
⇒ u = 1 × 1011 × 25 × 10-6 = 25 × 105 = 25
∴ Elastic potential energy density u = 25 J/m3.
If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ____ m.
Correct answer
Why this answer is correct
Calculation:
θ1 = sin-1(2λ/a)
θ2 = sin-1(3λ/a)
It is given that:
θ1 + θ2 = 30° = π/6
⇒ sin-1 (2λ/a) + sin-1(3λ/a) = π/6
using sin-1(x) + sin-1(y) = sin-1[x√(1-y2) +y√(1-x2)]
⇒ (2λ/a) × √(1 - (3λ/a)2) + (3λ/a) × √(1 - (2λ/a)2) = 1/2
Now plug in λ = 628 × 10-9 m and solve numerically to find a.
⇒ a ≈ 6.07 µm
is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If , then the value of n is ______.
Correct answer
Why this answer is correct
Calculation:
We are given a ratio of adiabatic indices (γ):
γA / γB = [(fA + 2) / fA] × [fB / (fB + 2)]
Substitute values:
= (3 + 2)/3 × (6 + 2)/(6 + 2 + 2)
= (5/3) × (8/10)
= 40 / 30
So:
40 / 30 = 1 + (1/n)
⇒ (40 / 30) - 1 = 1/n
⇒ 10 / 30 = 1/n
⇒ n = 3
A person travelling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocity v2 for the next distance. The average velocity in this motion is m/s. If v1 is 5 m/s then v2 = _____ m/s.
Correct answer
Why this answer is correct
Calculation:
The average velocity is given by:
vavg = (x1 + x2) / (t1 + t2)
Given: vavg = 50 / 7
Let x be the initial distance:
x1 = x,
x2 = (3x / 2),
t1 = x / 5,
and t2 = (3x / 2) / v2
⇒ 50 / 7 = (x + 3x / 2) / (x / 5 + (3x / 2v2))
⇒ 50 / 7 = (5 / 2) / (1 / 5 + 3 / (2v2))
⇒3 / (2v2) = 7 / 20 - 1 / 5 = (7 - 4) / 20 = 3 / 20
⇒3 / (2v2) = 3 / 20
⇒ v2 = 10 m/s
75 questions in the full paper