Physics Test

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Physics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

A ball of mass $m_{1}$ is moving with velocity $3 v$. It collides head on elastically with a stationary ball of mass $m_{2}$. The velocity of both the balls become $v$ after collision. Then the value of the ratio $\frac{m_{2}}{m_{1}}$ is

A
1

B
2

C
3

D
4

Solution

\[
\begin{array}{l}
3 m_{1} v=m_{1} v+m_{2} v \\
3 m_{1} v-m_{1} v=m_{2} v \\
2 m_{1} v=m_{2} v \\
\therefore \frac{m_{2}}{m_{1}}=2
\end{array}
\]
Question 2
1 / -0

A transparent plastic bag filled with air forms a concave lens. Now, if this bag is completely immersed in water, then it behaves as

A
Divergent lens

B
Convergent lens

C
Equilateral prism

D
Rectangular slab

Solution

Answer(B)

Convergent lens
Question 3
1 / -0

A lead bullet of 10g travelling at300 ms^-1strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = 150jkg^-¹k^-1)

A
100 $℃$

B
125 $℃$

C
150 $℃$

D
200 $℃$

Solution

ANSWER
Energy absorbed by the bullet=0.6 $\times$ (Kinetic energy of the bullet) $m c \Delta T=0.6(0.5) m v^{2}$
$\Delta T=\frac{(0.5)(0.5)\left(300^{2}\right)}{150}=150^{\circ} \mathrm{C}$
Question 4
1 / -0

A particle of mass M at rest decays into two masses m₁ and m₂with non-zero velocities. The ratio of de-Broglie wavelengths of the particles $\frac{λ_{1/}}{λ_{2}}$ is

A

B

C

D
1 : 1

Solution
Question 5
1 / -0

A sample of HCl gas is placed in an electric field of 3 × 10⁴NC^-1. The dipole moment of each HCl molecule is 6 × 10^-³⁰cm. The maximum torque that can act on a molecule is

A
2×10^-34 C² m N^-1

B
2×10^-34 N m

C
18×10^-26 N m

D
0.5×10³⁴C^-2 N m^-1

Solution

Torque,

\[
\begin{array}{l}
\vec{\tau}=\vec{p} \times \vec{E} \\
\tau_{\max }=p E=6 \times 10^{-30} \times 3 \times 10^{4} \\
=18 \times 10^{-26} \mathrm{Nm}
\end{array}
\]
Question 6
1 / -0

A particleA has a charge and particleB has charge 4 with each of them having the mass m. When they are allowed to fall from rest through same potential difference, the ratio of their speeds will be

A
4 : 1

B
1 : 4

C
1 : 2

D
2 : 1

Solution

Speed obtained by the particle after falling through a potential difference of V volt is
Question 7
1 / -0

If in Young's double slit experiment, d = 0.1 mm, D = 20 cm, and $λ = 5460 Å$ , then angular position of first dark fringe will be

A

B

C

D

Solution
Question 8
1 / -0

The adiabatic modulus of elasticity of gas is 2.1 × 10⁵ N m^-2. What will be its isothermal modulus of elasticity? (C_p/C_v=1.4)

A
1.2×10⁵ N m^-2

B
4×10⁵ N m^-2

C
1.5×10⁵N m^-2

D
1.8×10⁵ N m^-2

Solution

$\frac{E_{s}}{E_{T}}=\gamma=\frac{C_{P}}{C_{V}}=1.4$
$\frac{2.1 \times 10^{5}}{E_{T}}=1.4$
$=1.5 \times 10^{5} \mathrm{Nm}^{-2}=2 . \frac{2.1 \times 10^{5}}{1.4}$
$=\mathrm{or} \quad E_{T}=$
$=1.5 \times 10^{5}$

Adiabatic Bulk Modulus Adiabatic elasticity $\left(E_{\phi}\right): \mathrm{PV}^{\gamma}=$ constant
Differentiating both sides $\mathrm{d} \mathrm{PV}^{\gamma}+\mathrm{P} \gamma \mathrm{V}^{\gamma-1} \mathrm{dV}=0$
\[
\gamma \mathbf{P}=\frac{\mathrm{dP}}{-\frac{\mathrm{dv}}{\mathrm{v}}}=\frac{\text { Stress }}{\text { Strain }}=\mathrm{E}_{\phi} \Rightarrow \mathrm{E}_{\phi}=\gamma \mathrm{P}
\]
i.e., adiabatic elasticity is $\gamma$ times that of pressure
Also isothermal elasticity $\mathrm{E}_{\theta}=\mathrm{P} \Rightarrow \frac{\mathrm{E}_{\phi}}{\mathrm{E}_{\theta}}=\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$
i.e. the ratio of two elasticity of gases is equal to the ratio of two specific
Isothermal Bulk Modulus Isothermal elasticity $\left(\mathrm{E}_{\theta}\right):$ For this process $\mathrm{PV}=$ constant
\[
\Rightarrow P d V=-V d P \Rightarrow P=\frac{d P}{-\frac{d v}{v}}=\frac{\text { Stress }}{\text { Strain }}=E_{\theta}
\]
$\Rightarrow \mathrm{E}_{\theta}=\mathrm{P}$ i.e., isothermal elasticity is equal to pressure
At N.T.P $\mathrm{E}_{\theta}=$ Atmospheric pressure $=1.01 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}$
Question 9
1 / -0

A square coil ACDE with its plane vertical is released from rest in a horizontal uniform magnetic field of length 2L. The acceleration of the coil is

A
Less than g for all the time till the loop crosses the magnetic field completely

B
Less than g when it enters the field and greater than g when it comes out of the field

C
g all the time

D
Less than g when enters and comes out of the field but equal to g when it is within the field

Solution

Answer(D)

Less than g when enters and comes out of the field but equal to g when it is within the field
Question 10
1 / -0

The reverse voltage at which the current increases enormously, in a p-n junction, is called

A
Knee voltage

B
Breakdown voltage

C
Biasing voltage

D
Accelerating voltage

Solution

The reverse voltage at which the current increases enormously, in a p-n junction, is called Breakdown voltage.