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**Question 1: If the kinetic energy of a particle becomes four times, then % change in momentum will be:**

a. 200

b. 100

c. 150

d. 50

**Answer: (b)**

**\(\begin{array}{l}\frac{P_{2}}{P_{1}} = \sqrt{\frac{K_{2}}{K_{1}}} \Rightarrow \frac{P_{2}}{P_{1}} = \sqrt{\frac{4K}{K}}\end{array} \)
\(\begin{array}{l}\frac{P_{2}}{P_{1}} = 2\end{array} \)
\(\begin{array}{l}\frac{P_{2} – P_{1}}{P_{1}}% = \frac{P_{2}}{P_{1}-1}\times 100 = (2-1)\times 100 = 100\end{array} \)
\(\begin{array}{l}\frac{\Delta P}{P_{1}}% = 100%\end{array} \)
**

K.E. ⇒ K = P^{2} / 2m

P ∝ √K

⇒

⇒

⇒

**
**

**Question 2: A RLC circuit is in its resonance condition. Its circuit components have value R = 5Ω, L = 2H, C = 0.5mF, V= 250V. Then find the power in the circuit.**

a. 6 kW

b. 10 kW

c. 12 kW

d. 12.5 kW

**Answer: (d)**

**\(\begin{array}{l}P = i_{rms}^{2}R\end{array} \)
\(\begin{array}{l}P = \frac{V^{2}}{R} = \frac{250\times 250}{5} = 12500\frac{J}{S} = 12.5\: kW\end{array} \)
**

As the circuit is in resonance. Thus,

X_{L} = X_{C}

∴ Z = R so, i_{rms} = V / Z = V / R

**
****Question 3: A satellite is revolving around a planet in an orbit of radius R. Suddenly radius of orbit becomes 1.02 R, then what will be the percentage change in its time period of revolution?**

a. 2%

b. 3%

c. 5%

d. 7%

**Answer: (b)**

**\(\begin{array}{l}T \alpha R^{\frac{3}{2}}\end{array} \)
\(\begin{array}{l}T = kR^{\frac{3}{2}}\end{array} \)
\(\begin{array}{l}\frac{\Delta T}{T} = \frac{3}{2}\frac{\Delta R}{R} = 3%\end{array} \)%
**

**
\(\begin{array}{l}\frac{t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)**

b.\(\begin{array}{l}\frac{t_{1}t_{2}}{t_{1}-t_{2}}\end{array} \)

c.\(\begin{array}{l}\frac{2t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)

d.\(\begin{array}{l}\frac{2t_{1}t_{2}}{t_{1}-t_{2}}\end{array} \)

**Question 4: A person walks up a stationary escalator in the time 𝑡 _{1}. If he remains stationary on the escalator, then it can take him up in time 𝑡_{2}. Determine the time it would take to walk upon the moving escalator?**

a.

b.

c.

d.

**Answer: (a)**

**\(\begin{array}{l}\frac{L}{\frac{L}{t_{1}}+\frac{L}{t_{2}}} = \frac{t_{1}t_{2}}{t_{1}+t_{2}}\end{array} \)
**

Suppose length of escalator = L

Speed of man w.r.t escalator L / t_{1}

Speed of escalator = L / t_{2}

Speed of man w.r.t ground when escalator is moving = L / t_{1} + L / t_{2}

Time taken by the man to walk on the moving escalator =

**
****Question 5: For the given graph between decay rate and time find half-life (where R = decay rate).**

a. 10 / 3 ln2

b. 20 / 3 ln2

c. 3 / 20 ln2

d. 3 / 10 ln2

**Answer: (a)**

**\(\begin{array}{l}R = R_{o}e^{-\lambda t}\end{array} \)
\(\begin{array}{l}t_{1/2} = \frac{ln2}{\lambda} = \frac{ln2}{3}\times 20 = \frac{20}{3}ln2\end{array} \)
**

ln R = ln R_{o} – λt

Slope = -λ = – 6 / 40

λ = 3 / 20

**
****Question 6. A wheel rotating with an angular speed of 600 rpm is given a constant acceleration of 1800 rpm ^{2} for 10 sec. The number of revolutions revolved by the wheel is.**

a. 125

b. 100

c. 75

d. 50

**Answer: (a)**

ω_{o} = 600 rpm

α = 1800 rpm^{2}

t = 10 sec = 1/6 min

θ = ω_{o}t + ½ αt^{2}

θ = 600 ×10 / 60 + ½ × 1800 × 1 / 36

θ = 100 + 25 = 125 revolutions

**
****Question 7: **

**\(\begin{array}{l}\left | \vec{P} \right |=\left | \vec{Q} \right |,\left | \vec{P}+\vec{Q} \right |=\left | \vec{P}-\vec{Q} \right |\end{array} \)**

**. Find the angle between\(\begin{array}{l}\left | \vec{P} \right |\end{array} \) and \(\begin{array}{l}\left | \vec{Q} \right |\end{array} \)**

a. 45°

b. 90°

c. 135°

d. 150°

**Answer: (b)**

**\(\begin{array}{l}\left | \vec{P} + \vec{Q}\right | = \left | \vec{P} – \vec{Q}\right |\end{array} \)
\(\begin{array}{l}\left | \vec{P} \right |^{2} + \left | \vec{Q}\right |^{2} + 2\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta = \left | \vec{P} \right |^{2} + \left | \vec{Q}\right |^{2} – 2\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta\end{array} \)
\(\begin{array}{l}\left | \vec{P} \right |\left | \vec{Q}\right | \: cos\Theta = 0\end{array} \)
**

θ = 90°

**
****Question 8: Time (T), Velocity (C) and angular momentum (h) is chosen as fundamental quantities instead of mass, length and time. In terms of these, dimension of mass would be**

a. [M] = [T^{-1} C^{-2} h]

b. [M] = [T^{-1} C^{2} h]

c. [M] = [T^{-1} C^{-2} h^{-1} ]

d. [M] = [TC^{-2} h]

**Answer: (a)**

M ∝ T^{x} c^{y} h^{z}

^{1}L^{0}T^{0}] = [T^{x}] [LT^{-1}]^{y}[ML2T^{-1}]^{z}On comparing the powers, we get:

Z = 1 …………. (1)

X – y + z = 0. ……….(2)

Y +2z = 0 …………. (3)

So, y = -2

X = -1

[M] = [T^{-1}C^{-2}h]

**
****Question 9: Find the relation between 𝛾 (adiabatic constant) and degree of freedom (𝑓). **

a. f = 2 / γ – 1

b. f = γ / γ – 1

c. f = γ – 1 /2

d. f = γ-1 / γ

**Answer: (a)**

We know that,

C_{v} = *f* R / 2

C_{p} = (f / 2 + 1)*R* R

γ = C_{p} / C_{v} = 1 + 2 / f

F = 2 / γ – 1

**
****Question 10: Two identical drops of Hg coalesce to form a bigger drop. Find the ratio of the surface energy of a bigger drop to a smaller drop. **

a. 2 ^{3/2 }

b. 3^{ 2/5}

c. 2 ^{2/3}

d. 5 ^{2/3}

**Answer: (c)**

**\(\begin{array}{l}2\times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3}\end{array} \)
\(\begin{array}{l}\frac{R}{r} = 2^{\frac{1}{3}}\end{array} \)
\(\begin{array}{l}\frac{U_{bigger}}{U_{smaller}}=\frac{S\times4\pi R^{2}}{S\times4\pi r^{2}}=(\frac{R}{r})^{2} = 2^{\frac{2}{3}}\end{array} \)
**

**
\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}+x^{\frac{2}{1}}}{v^{\frac{2}{1}}-v^{\frac{2}{2}}}}\end{array} \)**

b.\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}-x^{\frac{2}{1}}}{v^{\frac{2}{1}}+v^{\frac{2}{2}}}}\end{array} \)

c.\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}-x^{\frac{2}{1}}}{v^{\frac{2}{1}}-v^{\frac{2}{2}}}}\end{array} \)

d.\(\begin{array}{l}2\pi \sqrt{\frac{x^{\frac{2}{2}}+x^{\frac{2}{1}}}{v^{\frac{2}{1}}+v^{\frac{2}{2}}}}\end{array} \)

**Question 11: The velocities of a particle performing SHM at a distance of x _{1} and x_{2} from mean position are v_{1} and v_{2}, find the time period of oscillation?**

a.

b.

c.

d.

**Answer: (c)**

**\(\begin{array}{l}v = \omega \sqrt{A^{2} – x^{2}}\end{array} \)
\(\begin{array}{l}v_{1} = \omega \sqrt{A^{2} – x_{1}^{2}}\end{array} \)
\(\begin{array}{l}v_{2} = \omega \sqrt{A^{2} – x_{2}^{2}}\end{array} \)
\(\begin{array}{l}(\frac{v_{1}}{\omega})^{2} – (\frac{v_{2}}{\omega})^{2} = x_{2}^{2} – x_{1}^{2}\end{array} \)
\(\begin{array}{l}\omega^{2} = \frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2} – x_{1}^{2}}\end{array} \)
\(\begin{array}{l}\omega = \sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2} – x_{1}^{2}}}\end{array} \)
\(\begin{array}{l}T =2\pi \sqrt{\frac{x_{2}^{2} – x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}\end{array} \)
**

**
****Question 12: Identify the correct graph between PV and T for an ideal gas.**

**Answer: (c)**

PV = nRT ⇒ PV=CT

Therefore, PV v/s T graph is a straight line.

**
****Question 13: In photoelectric effect stopping potential is 3V _{0} for incident wavelength λ_{0} and stopping potential for incident wavelength 2λ_{0}. Find threshold wavelength.**

a. 3λ_{0 }

b. 2λ_{0}

c. 4λ_{0}

d. 8λ

**Answer: (c)**

KE = hv – W

eV = hc / λ – W

For first case

e(3V_{o}) = hc /λ_{o} – W ……..(i)

For second case:

e(V_{o})=hc / 2λ_{o} – W ……(ii)

From equation (i) and (ii), we get,

W = hc / (4λ_{o})

For λ_{th}

w = hc / λ_{th}

⇒ hc / (4λ_{o}) = hc/λ_{th}

⇒ λ_{th} = 4λ_{o}

**
\(\begin{array}{l}\hat{j}\end{array} \)**

b.\(\begin{array}{l}-\hat{k}\end{array} \)

c.\(\begin{array}{l}-\hat{j}\end{array} \)

d.\(\begin{array}{l}\hat{k}\end{array} \)

**Question 14: A plane electromagnetic wave travels in free space. Electric field is **

**\(\begin{array}{l}\left | \vec{E} \right |=E_{0}\hat{i}\end{array} \)**

**and magnetic field is represented by\(\begin{array}{l}\left | \vec{B} \right |=B_{0}\hat{k}\end{array} \). What is the unit vector along the direction of propagation of electromagnetic waves?**

a.

b.

c.

d.

**Answer: (c)**

**\(\begin{array}{l}\vec{E}\times \vec{B}\end{array} \).
\(\begin{array}{l}\vec{E}\times \vec{B}\end{array} \), = \(\begin{array}{l}\frac{\vec{E}\times \vec{B}}{\left | \vec{E}\times \vec{B} \right |}\end{array} \)
\(\begin{array}{l}\frac{E_{o}\hat{i}\times B_{o}\hat{k}}{E_{o}B_{o}Sin\: 90^{o}}\end{array} \)
\(\begin{array}{l}\hat{i}\times \hat{k}\end{array} \)
\(\begin{array}{l}-\hat{J}\end{array} \)
**

Direction of the EM wave is given by direction of

Unit vector in direction

**
****Question 15: Two satellites of mass M _{A} and M_{B} are revolving around a planet of mass M in radius R_{A} and R_{B} respectively. Then**

a. T_{A }> T_{B}if R_{A} > R_{B }

b. T_{A}> T_{B} if M_{A}> M_{B}

c. T_{A}= T_{B} if M_{A}> M_{B}

d. T_{A} > T_{B} if R_{A}< R_{B}

**Answer: (a)**

**\(\begin{array}{l}\frac{T_{A}}{T_{B}} = \left ( \frac{R_{A}}{R_{B}} \right )^{\frac{3}{2}}\end{array} \)
**

According to Kepler’s law of planetary motion,

T ∝ R^{3/2}

So, if R_{A} > R_{B} then T_{A} > T_{B}.

Hence, option (a) is the correct answer.

**
****Question 16: At 45° of the magnetic meridian angle of dip is 30° then find the angle of dip in a vertical plane at 45°?**

a. tan^{−1}(1/√6)

b. tan^{−1}(1/√2)

c. tan^{−1} (1/√4)

d. tan^{−1}(1/√3)

**Answer: (a)**

Let the horizontal and vertical components of the earth’s magnetic field at the meridian be V and H.

Angle of dip, tan θ = V / H……(i)

At angle of 45° from magnetic meridian, angle of dip = 30°

tan 30° = V / H cos 45°

⇒1 / √3 = V / H cos 45°

V / H = 1 / √6

tan θ = V / H = 1 / √6

⇒θ = tan^{-1} (1 / √6)

**
****Question 17: A sodium lamp in space was emitting waves of wavelength 2880 Å. When observed from a planet, its wavelength was recorded at 2886 Å. Find the speed of the planet?**

a. 4.25 × 10^{5} m/s

b. 6.25 × 10^{5} m/s

c. 2.75 × 10^{5} m/s

d. 3.75 × 10^{5} m/s

**Answer: (b)**

**\(\begin{array}{l}\frac{V_{rel}}{C} = \frac{\Delta \lambda}{\lambda}\end{array} \)
\(\begin{array}{l}V_{rel} = \frac{6}{2880}\times3 \times 10^{8}\end{array} \)
\(\begin{array}{l}V_{rel} = 6.25 \times 10^{5} m/s\end{array} \)
**

**
****Question 18: For a body in pure rolling, its rotational kinetic energy is 1/2 times of its translational kinetic energy. The body should be? **

a. Solid cylinder

b. Ring

c. Solid sphere

d. Hollow sphere

**Answer: (a)**

Given,

Rotational K.E = ½ Translational K.E

½ Iω^{2} = ½ × ½ mv^{2}

In pure rolling, v = Rω

½ Iω^{2} = ¼ mR^{2} ω^{2}

I = ½ mR^{2}

Hence, it is a solid cylinder.

**
****Question 19: Magnetic susceptibility of a material is 499 & μ _{0} = 4π × 10^{−7} SI unit. Then find μ_{r}**

a. 500

b. 400

c. 300

d. 200

**Answer: (a)**

Given, 𝜒 = 499

The relative permeability, μ_{r} = 1 + 𝜒 = 500

**
****Question 20: An electron having the de Broglie wavelength λ falls on an X-ray tube. The cut-off wavelength of emitted X-ray is**

a. 2mcλ^{2 }/ h

b. 2h / mc

c. h / mc

d. 2mcλ^{2 }/ 3h

**Answer: (a)**

De-broglie wavelength, λ_{B} = h / P

P = h / (λ_{B})

Kinetic energy of electron, E = P^{2} / (2m_{e}) = h^{2} / 2m_{e} λ_{B}^{2}

For cutoff wavelength of emitted X-ray: E = hc / λ

h^{2} / 2m_{e} λ_{B}^{2 }= hc / λ

λ = 2m_{e} λ_{B}^{2} / h = 2mc λ^{2 }/ h where λ_{B} = λ and m_{e} = m

**
****Question 21: A body is moved from rest along a straight line by a machine delivering constant power. Time taken by the body to travel a distance 𝑆 is proportional to:**

a. 𝑆^{1/3}

b. 𝑆^{2/3}

c. 𝑆^{1/2}

d. 𝑆^{1/4}

**Answer: (b)**

**\(\begin{array}{l}\int_{O}^{S}ds = C \int_{O}^{t} t^{\frac{1}{2}}dt\end{array} \)
\(\begin{array}{l}t^{\frac{3}{2}} = \frac{3S}{2C} \Rightarrow t = S^{\frac{2}{3}}\left ( \frac{3}{2C} \right ) \Rightarrow T\: \alpha \: S^{\frac{2}{3}}\end{array} \)
**

Energy supplied in time t sec is,

E = P × t

Here P represents the power delivered by the machine.

⇒ Pt = ½ mv^{2}

v ∝ √t

Writing the velocity in terms of the displacement of the body,

⇒ dS/dt = C√t

Here C is a constant.

⇒

**
\(\begin{array}{l}\frac{L_{1}T_{1}-L_{2}T_{2}}{T_{1}-T_{2}}\end{array} \)**

b.\(\begin{array}{l}\frac{L_{2}T_{1}-L_{1}T_{2}}{T_{2}-T_{1}}\end{array} \)

c.\(\begin{array}{l}\frac{L_{1}T_{2}-L_{2}T_{2}}{T_{2}-T_{1}}\end{array} \)

d.\(\begin{array}{l}\frac{L_{1}}{T_{1}}\times \frac{T_{2}}{L_{2}}\end{array} \)

**Question 22: A uniform rod of Young’s modulus Y is stretched by two tension forces T _{1} and T_{2} such that the rods get expanded to length L_{1} and L_{2} respectively. Find the initial length of the rod?**

a.

b.

c.

d.

**Answer: (c)**

**\(\begin{array}{l}\frac{F}{A} = Y\frac{\Delta l}{l}\end{array} \)
\(\begin{array}{l}\frac{T}{A} = Y\frac{\Delta l}{l}\end{array} \)
\(\begin{array}{l}\frac{T_{1}}{A} = \frac{Y(L_{1}-L_{o})}{L_{o}}\end{array} \) ……(i)
\(\begin{array}{l}\frac{T_{2}}{A} = \frac{Y(L_{2}-L_{o})}{L_{o}}\end{array} \)……(ii)
\(\begin{array}{l}\frac{T_{1}}{T_{2}} = \frac{L_{1}-L_{o}}{L_{2}-L_{o}}\end{array} \)
\(\begin{array}{l}L_{o} = \frac{L_{1}T_{2}-L_{2}T_{1}}{T_{2}-T_{1}}\end{array} \)
**

Let the initial length of the rod to be L_{o} and area A

In this case the external force on the rod is tension hence from Hook’s law,

⇒

For both the cases we can relate the tension and elongation produced in rod as,

Here (L_{1} – L_{o}) is the elongation in rod when T_{1} is applied.

Similarly,

Here (L_{2} – L_{o}) is the elongation in rod when T_{2} is applied.

On dividing Eq.(i) and (ii) we get,

⇒ T_{1} L_{2} – T_{1} L_{o} = T_{2} L_{1} – T_{2} L_{o}

Or,

**Question 23: A block is projected up to a rough plane of inclination 30°. If time of ascending is half the time for descending and the coefficient of friction is μ =3/5√𝑛. Then 𝑛 =**

**Answer: (n = 3)**

**\(\begin{array}{l}\frac{t_{A}^{2}}{t_{D}^{2}} = \frac{a_{D}}{a_{A}}\end{array} \)
\(\begin{array}{l}\frac{t_{A}^{2}}{t_{D}^{2}} = \frac{g \: sin \: \Theta – \mu g \: cos \Theta}{g \: sin \: \Theta + \mu g \: cos \Theta}\end{array} \)
\(\begin{array}{l}\frac{t_{A}}{t_{D}} = \sqrt{\frac{g \: sin \: \Theta – \mu g \: cos \Theta}{g \: sin \: \Theta + \mu g \: cos \Theta}}\end{array} \)
\(\begin{array}{l}\frac{1}{2} = \sqrt{\frac{1-\sqrt{3\mu }}{1+\sqrt{3\mu }}}\end{array} \)
**

S = ½ a_{A}t_{A}^{2} ….. (1)

S = ½ a_{D}t_{D}^{2} ….. (2)

From equation (1) and (2)

⇒

⇒

⇒

⇒ 1 + √3μ = 4 – 4√3 μ

⇒ 5√3μ = 3

⇒ μ = 3 / 5√3

**Question 24: I-V characteristic curve of a diode in a forward bias is given in fig. Find out dynamic resistance**

a. 212.3 Ω

b. 205.3 Ω

c. 245.3 Ω

d. 233.3 Ω

**Answer: (d)**

Dynamic resistance = ΔV / ΔI

= 0.7 V / 3 mA = 233.3 Ω

**
****Question 25: An electron is accelerated through a voltage of 40 kV. What will be its wavelength?**

a. 0.061 Å

b. 0.011 Å

c. 0.021 Å

d. 0.161 Å

**Answer: (a)**

**\(\begin{array}{l}\frac{12.27}{\sqrt{V}}\end{array} \) Å
\(\begin{array}{l}\frac{12.27}{\sqrt{40 \times 10^{3}}}\end{array} \) = 0.061 Å
**

λ_{B} = h / P

= h / √2meV

=

=

**Question 26: Find the value of R _{p} in the given circuit? (V_{Z} = 8V)**

a. 4Ω

b. 6Ω

c. 3Ω

d. 5Ω

**Answer: (b)**

Applying KVL

20 – 8 – 2R_{P} = 0

R_{P} = 6 Ω

**
\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}-r_{2})^{3}}{G(m_{1}+m_{2})}}\end{array} \)**

b.\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}+r_{2})^{3}}{G(m_{1}+m_{2})}}\end{array} \)

c.\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}-r_{2})^{3}}{G(m_{1}-m_{2})}}\end{array} \)

d.\(\begin{array}{l}2\pi \sqrt{\frac{(r_{1}+r_{2})^{3}}{G(m_{1}-m_{2})}}\end{array} \)

**Question 27: Two stars of masses m _{1} and m_{2} are in mutual interaction and revolving in orbits of radii r_{1} and r_{2} respectively. The time period of revolution for this system will be?**

a.

b.

c.

d.

**Answer: (b)**

**\(\begin{array}{l}\frac{Gm_{1}m_{2}}{(r_{1}+r_{2})^{2}} = m_{1}r_{1}\omega ^{2} = m_{1}\times \frac{m_{2}(r_{1}+r_{2})}{m_{1}+m_{2}}\omega ^{2}\end{array} \)
\(\begin{array}{l}\omega = \frac{\sqrt{G(m_{1}+m_{2})}}{(r_{1}+r_{2})^{\frac{3}{2}}}\end{array} \)
\(\begin{array}{l}T = \frac{2\pi }{\omega }\end{array} \)
\(\begin{array}{l}T = 2\pi\frac{(r_{1}+r_{2})^{3}}{G(m_{1}+m_{2})}\end{array} \)
**

Let angular velocity will be ω

For mass m_{1},

**
****Question 28: If N _{0} active nuclei become N_{0}/16 in 80 days. Find half-life of nuclei?**

a. 40 days

b. 20 days

c. 60 days

d. 30 days

**Answer: (b)**

**
****Question 29: A gas is undergoing a change in state by an isothermal process AB as follow. Work done by gas in process AB is**

a. 100 ln2 Joule

b. −100 ln2 Joule

c. 200 ln2 Joule

d. −200 ln2 Joule

**Answer: (c)**

W_{isothermal} = P_{1}V_{1} ln V_{2} / V_{1}

V_{1} = 100 m^{3}

V_{2} = 200 m^{3}

P_{1} = 2 N/m^{2}

W_{isothermal} = 2 × 100 ln 200 / 100

W = 200 ln 2 Joule

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