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Given:

 \(p(\hat{i}+\hat{j}+\hat{k})\) is a unit vector.

Concept:

A unit vector of a vector is a vector of unit magnitude and direction as that of the original vector.

Formula:

The magnitude of a vector aî + bĵ + ck̂ is given by √(a2 + b2 +c2)

Calculation:

The magnitude of given vector  \(p(\hat{i}+\hat{j}+\hat{k})\) can be calculated as;

| \(p(\hat{i}+\hat{j}+\hat{k})\)|= |p|√(1² + 12 + 12)

= |p|√3

∵ it is a unit vector:

⇒ |p|√3 = 1

⇒ |p| = 1/√3

or p = ±1/√3

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as \(\vec a \cdot \left( {\vec b \times ;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\ {{b_1}}&{{b_2}}&{{b_3}}\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a;b;c} \right]\)

Calculation:

Given: \(\vec a = \hat i + \hat j + \hat k;,;\vec b = 6\hat i + 6\hat j + 6\hat k;and;\vec c = 2\hat i + 3\hat j + \hat k\)

As we know that, the scalar triple product of three vectors is \(\vec a \cdot \left( {\vec b \times ;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\ {{b_1}}&{{b_2}}&{{b_3}}\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a;b;c} \right]\)

⇒ \(\vec a \cdot \left( {\vec b \times ;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{1}}&{{1}}&{{1}}\ {{6}}&{{6}}&{{6}}\ {{2}}&{{3}}&{{1}} \end{array}} \right|\)

⇒ \(\vec a \cdot \left( {\vec b \times ;\vec c} \right) = 1 \times (6 - 18) - 1 \times (6 - 12) + 1 \times (18 - 12) = 0\)

Hence, the correct option is 2.

Calculation

All given equation of circle in option have same center at \((0,3)\).

Now,  eccentricity of ellipse,e \(=\sqrt {1-\dfrac {b^2}{a^2}}\)

⇒ e \(=\sqrt{1-(9/16)}=\sqrt {7}/4 \)

\(⇒ \) foci of ellipse \(=(\pm ae,0)=(4\times \sqrt{7}/4, 0)=(\pm\sqrt{7},0)\)

The center of the circle is \((0,3)\)

Radius \(=\sqrt{(\sqrt{7}-0)^{2}+(0-3)^{2}}=\sqrt{7+9} =4 \).

Hence, the equation of the circle is :

⇒ \((x-0)^2 +(y-3)^2 = 4^2 =16 \)

⇒ \(x^2 +y^2-6y-7=0\)

Hence option 4 is correct

Calculation: 

By squeeze theorem-

Since, ∀x > 0, 0 ≤ \(\rm\left|\frac{\sin x}{x}\right| ≤\left|\frac{1}{x}\right|\), ∀x ≥ \(\rm\frac{1}{x}\)

and since, \(\rm\displaystyle\lim_{x \rightarrow \infty} \frac{1}{x}\) = 0

⇒ \(\rm\displaystyle\lim_{x \rightarrow \infty} \sin x \times \frac{1}{x} = \lim_{x \rightarrow \infty} \frac{\sin x}{x}\) = 0.

Alternate solution-

y = sin x, x ∈ R

−1 ≤ sin x ≤ 1

\(\rm\displaystyle\lim_{x \rightarrow \infty} \frac{\sin x}{x} = \frac{\sin \infty}{\infty} = \frac{[−1, 1]}{\infty}\) = 0

The correct answer is option "2"

Concept:

\( \rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

To Find: Integral of sec² x with respect to sec x

\(\rm \int \sec^2 x; d(\sec x)\)

Replace sec x = t, we get

\(= \rm \int t^2 dt\= \frac{t^3}{3}+c\=\frac{\sec^3 x}{3}+c\)

CONCEPT:

• Percentage error is the difference between theoretical value and an experimental, divided by the theoretical value, multiplied by 100 to give a percent.
• To calculate percentage error in the expression A = xmyn / zp . We use formula

\(\frac{{\Delta A}}{A} × 100 = m\frac{{\Delta x}}{x}× 100 + n\frac{{\Delta y}}{y} × 100+ p\frac{{\Delta z}}{z}× 100\)

where \(\frac{{\Delta A}}{A} × 100\) is the percentage error in A and \(\frac{{\Delta x}}{x} × 100\) is the percentage error in x.

• The Kinetic Energy (E) of an object due to its linear speed is given by:

​​E = 1/2 (m × v2) 

where m is the mass of a body and v is the speed.

CALCULATION:

Given in the question:

\(\frac{{\Delta m}}{m}× 100 = 1%\)

\(\frac{{\Delta v}}{v}× 100 = 2%\)

Now we have to find the error in ​​E = 1/2 (m × v2) 

\(\frac{{\Delta E}}{E} × 100 = \frac{{\Delta M}}{M}× 100 + 2\frac{{\Delta v}}{v} × 100\)

\(\frac{{\Delta E}}{E} × 100 =1 + 2\times 2=5 %\)

∴ % error in the estimation of the kinetic energy of the body is = 5%

• So the correct answer will option 4.

CONCEPT:

• Force: The interaction which after applying on a body changes or try to change the state of rest or state of motion of the body is called force.

Force (F) = Mass (m) × acceleration (a) 

• Retardation: The force acting in the opposite direction to the motion of a body is called the retarding force. This force produces negative acceleration for the body which is called retardation or deacceleration.

Retardation (a) = Force/Mass

• The equation of motion is given below:

V = u + a t

Where V is final velocity, u is initial velocity, a is acceleration and t is time

CALCULATION:

Given: m = 20 kg, u = 15 m/s, F = - 50 N, v = 0 m/s​

• The acceleration of the body under the retarding force F,

⇒ a = F/m = - 50/20 = - 2.5 m/s2

• We know that body is stopped, it means the final velocity is 0.
• Using equation 1,

⇒ 0 = 15 + (- 2.5)t

⇒ t = 6 sec

• Time taken by the body to stop will be 6 seconds.

So option 1 is correct.

Explanation:

Initially, the body has zero velocity and zero slope.

Thus,the acceleration would be zero initially.

Afterthat, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

Thus the correct option is 3)

Calculation:

Given: |P + Q| = |P| = |Q|

Let the magnitude of both vectors be P (since |P| = |Q|).

Now use the formula for magnitude of the sum of two vectors:

|P + Q|² = P² + Q² + 2PQ cosθ

Since |P + Q| = P, we have:

P² = P² + P² + 2P² cosθ

Simplifying:

P² = 2P² + 2P² cosθ

⇒ P² = P² (2 + 2 cosθ) 

⇒ 2 + 2 cosθ = 1

⇒ 2cosθ = -1

⇒ θ = 120°

Correct Answer: Option (2) → 120°

Calcultion:

F – mg = ma

F = ma + mg

F – (m – x)g = (m – x) 3a

Put F

ma + mg – mg + xg = 3ma – 3xa

\(x=\frac{2 m a}{g+3 a} \)