Physics Test

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Physics Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The average degree of freedom per molecule of a gas is 6. The gas performs 25 J work, while expanding at constant pressure. The heat absorbed by the gas is:-

A
75 J

B
100 J

C
150 J

D
125 J

Solution

Change in the internal energy of gas isss $\triangle U=n C_{v} \triangle T=\frac{f}{2} n R \triangle T$
For a constant pressure process. Work done $(\mathrm{W})=\mathrm{nR} \Delta \mathrm{T}$ So heat absorbed by the gas is $\triangle Q=\triangle U+W$
$\triangle Q=\frac{f}{2} n R \triangle T+n R \triangle T$
$\triangle Q=\frac{6}{2} n R \triangle T+n R \triangle T$
$\triangle Q=4 n R \triangle T$
$\triangle Q=4 \times 25=100 J$
Hence, the correct option is (B)
Question 2
1 / -0

All wires have same resistance and equivalent resistance between A and B is R. Now keys are closed, then the equivalent resistance will become:-

A
$\frac{7 R}{3}$

B
$\frac{7 R}{9}$

C
7R

D
$\frac{R}{9}$

Solution

$\Rightarrow \frac{7 R_{0}}{3}=R_{e q}$

As $R_{0}=R / 3$

$R_{e q}=7 R / 9$
Hence, the correct option is (B)
Question 3
1 / -0

A point charge +Q is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is:-

A
$\frac{Q}{16 \varepsilon_{0}}$

B
$\frac{Q}{4 \varepsilon_{0}}$

C
$\frac{Q}{8 \varepsilon_{0}}$

D
None of these

Solution

Flux going in pyramid $=\frac{Q}{2 \varepsilon_{0}}$ Which is divided equally among all 4 faces

$\therefore$ Flux through one face $\frac{Q}{8 \varepsilon_{0}}$
Hence, the correct option is (C)
Question 4
1 / -0

The radioactive sources A and B have half lives of 2hr and 4hr respectively, initially contain the same number of radioactive atoms. At the end of 2 hours, their rates of distintegration are in the ratio:-

A
4 : 1

B
2 : 1

C
$\sqrt{2} : 1$

D
1 : 1

Solution

Given that the half-life of radioactive source A is, t_1/2 = 2 hr
half-life of radioactive source B is, t_1/2 = 4 hr
Decay constant of radioactive source A is,

$\lambda_{A}=\frac{0.693}{t_{1 / 2}}$
$\lambda_{A}=\frac{0.693}{2}$
Decay constant of radioactive source $B$ is. $\lambda_{B}=\frac{0.693}{t_{1 / 2}}$
$\lambda_{B}=\frac{0.693}{4}$
The rate of disintegration (A) of source A at the end 2 hours is, $A_{A}=\frac{A_{O}}{t / t_{1 / 2}}$
$A_{A}=\frac{\lambda_{A} N_{A}}{2^{2 / 2}}$
$A_{A}=\frac{\lambda_{A} N_{A}}{2}$
The rate of disintegration (A) of source B at the end 2 hours is. $A_{B}=\frac{A_{o}}{t / t_{1 / 2}}$
$A_{B}=\frac{\lambda_{B} N_{B}}{2^{2 / 4}}$
$A_{B}=\frac{\lambda_{B} N_{B}}{\sqrt{2}}$
The ratio of rate of distintegration of source $A$ to $B$ iss $s$ $\frac{A_{A}}{A_{B}}=\frac{\lambda_{A} N_{A}}{2} \times \frac{\sqrt{2}}{\lambda_{B} N_{B}}$
$\frac{A_{A}}{A_{B}}=\frac{\lambda_{A}}{2} \times \frac{\sqrt{2}}{\lambda_{B}} \quad\left(N_{A}=N_{B}\right)$
$\frac{A_{A}}{A_{B}}=\frac{\lambda_{A}}{\lambda_{B} \sqrt{2}}$
$\frac{A_{A}}{A_{B}}=\frac{0.693}{2} \times \frac{4}{0.693 \sqrt{2}}$
$\frac{A_{A}}{A_{B}}=\sqrt{2}$
Hence, the correct option is (C)
Question 5
1 / -0

A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each having negligible resistance. There is a 2Ω resistor and 10V battery as shown in figure. There is a uniform magnetic field directed into the plane of the paper of magnitude 0.5 T. A force of 0.5N to the left is required to keep the wire PQ moving with constant speed to the right. With what speed is the wire PQ moving? (Neglect self-inductance of the loop)

A
8 m/s

B
16 m/s

C
24 m/s

D
32 m/s

Solution
Question 6
1 / -0

A charge particle q of mass m₀ is projected along the y-axis at t = 0 from origin with a velocity V₀. If a uniform electric field E₀ also exists along the x-axis, then the time at which the de-Broglie wavelength of the particle becomes half of the initial value is:

A

$\frac{m_{o} v_{o}}{q E_{o}}$

B
$\frac{2 m_{o} v_{o}}{q E_{o}}$

C
$\frac{\sqrt{3} m_{o} v_{o}}{q E_{o}}$

D
$\frac{3 m_{o} v_{o}}{q E_{o}}$

Solution

Force on charged particle is =qE₀ in x direction.
At any time t velocity in y direction will be same as at t=0 as there is no force in that direction.
At time t velocity in x direction is given by
v_x =a_xt
v_x =(qE₀/m)t

so speed of particle at time $t$ is- $\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{\left(\frac{q E_{0}}{m} t\right)^{2}+v^{2}}$
debroglie wavelength is inversly proportional to speed of particle and given by$\lambda_{0}=\frac{R}{m v_{0}}$ (Where $R$ is a constant $)$ So for half the wavelength speed should be doubled. $\frac{\lambda_{0}}{2} \Rightarrow v=2 v_{0}$
$v_{x}^{2}+v_{0}^{2}=4 v_{0}^{2}$
$v_{x}=\sqrt{3} v_{0}$
$\frac{q E_{0}}{m_{0}} t=\sqrt{3} v_{0}$
$t=\frac{\sqrt{3} v_{0} m_{0}}{q E_{0}}$
Hence, the correct option is (C)
Question 7
1 / -0

In the given figure a ring of mass m is kept on a horizontal surface while a body of equal mass 'm' attached through a string, which is wounded on the ring. When the system is released the ring rolls without slipping. Consider the following statements and choose the correct option.

(i) acceleration of the centre of mass of ring $i s \frac{g}{3}$
(ii) acceleration of the hanging particle is $\frac{2 g}{3}$
(iii) frictional force (on the ring) acts along forward direction
(iv) frictional force (on the ring) acts along backward direction

A
statement (i) and (ii) only

B
statement (ii) and (iii) only

C
statement (iii) and (iv) only

D
none of these

Solution

Free body diagram of ring and mass is

Since the ring is in pure rolling so its above point acceleration will be 2a and similarly acceleration of mass will be 2a.
The equation of motion of the system is
mg – T = m(2a) ......(1)
T = ma .......(2)
Substitute the value of T in equation (1)
mg - ma = 2ma
3ma = mg
a = g/3
So the acceleration of the center of the ring is, g/3
Andacceleration of the hanging mass is 2a = $\frac{2 g}{3}$
Hence, the correct option is (A)
Question 8
1 / -0

A carrier wave has power of 1675 kW. If the side band power of a modulated wave subjected to 60%. Then find the amplitude modulation level

A
301.5KW

B
942KW

C
1392KW

D
1884KW

Solution

As you know that, sideband power is, $P_{S}=\frac{1}{2} m^{2} P_{C}$, Where $P_{C}$ is power of carrier wave, $P_{S}$ is sideband power and 'm' is modulation rate, $P_{C}=\frac{2 P_{S}}{m^{2}}$
$P_{S}=\frac{m^{2} P_{C}}{2}$
$P_{S}=\frac{0.6 \times 0.6 \times 1675}{2}$
$P_{S}=301.5 \mathrm{kW}$
Hence, the correct option is (A)
Question 9
1 / -0

A telescope has an objective of focal length 60 cm and eyepiece of focal length 7 cm. The least distance of distinct vision is 35 cm. The telescope is focused for distinct vision on a scale 240 cm away from the objective. Find the produced magnification

A
0.23

B
0.56

C
0.75

D
0.80

Solution

For image $I_{1}$ of objective formed by objective lens, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{60}+\frac{1}{240}$
$\frac{1}{v}=\frac{5}{240}$
$v=48 \mathrm{cm}$
Magnification produced by the objective lens is $m_{o}=\frac{v}{u}$ $m_{o}=\frac{48}{240}=\frac{1}{5}$
Image $I_{1}$ acts as an object for eye lens. Hence $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$
$\frac{1}{u}=\frac{1}{35}-\frac{1}{7}$
$\frac{1}{u}=-\frac{4}{35}$
$u=-8.75$
Magnification produced by the eye lens is $m_{e}=\frac{v}{u}$ $m_{e}=\frac{35}{8.75}=4$
Produced magnification is
$m=m_{o} \times m_{e}$
$m=\frac{1}{5} \times 4$
$m=\frac{4}{5}$
$m=0.8$
Hence, the correct option is (D)
Question 10
1 / -0

In given P-V diagram, if the temperature of Point B is 25^oC, then find the ratio of temperature at point C and E.

A
3:1

B
1:3

C
2:5

D
5:2

Solution