Physics Test

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Physics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

A
3.6 ampere

B
2.8 ampere

C
2.5 ampere

D
5.0 ampere

Solution

A 220 volt input is supplied to a transformer.The output circuit draws a current of 2.0 ampere at
440 volts. If the efficiency of the transformer is $80 \%,$ the current drawn by the primary windings of the transformer is
$\eta=e f f i c i e n c y=\frac{\text {Output power}}{\text {Input Power}}=\frac{V_{s} I_{s}}{V_{p} I_{p}}$
$I_{p}=\frac{440 \times 2}{\frac{80}{100} \times 220}=5 A$
Hence, the correct option is (D)
Question 2
1 / -0

A cylindrical solid of length $1 m$ and radius $1 m$ is connected across a source of emf 10 V and negligible internal resistance shown in figure. The resistivity of the rod as a function of $x$ ( $x$ measured from left end) is given by $\rho=b x[$ where b is a positive constant]. Find the electric field (in SI unit) at point $P$ at a distance $10 cm$ from left end.

A
1

B
2

C
3

D
4

Solution

Resistivity of the rod $\rho=b x$
Area of rod $A=\pi r^{2}=\pi \times 1^{2}=\pi$
Total resistance of the cylinder $=\int_{0}^{1} \rho \frac{d x}{A}$
$=\int_{0}^{1} b \frac{x d x}{\pi}$
$=\frac{b}{\pi}\left(\frac{x^{2}}{2}\right)_{0}^{1}$
$=\frac{b}{\pi}\left[\frac{1}{2}-0\right]$
$R=\frac{b}{2 \pi}$
$E=-\frac{d v}{d x}=\frac{\left(i d R_{x}\right)}{d x}=i \frac{1}{d x}\left(b \frac{x d x}{A}\right) $
$=i\left(\frac{b x}{A}\right) $
$=i \rho \frac{x}{\pi} $
$E=\frac{10}{b / 2 \pi} \cdot B \frac{x}{\pi} $
$E=10 \times 2 \times x $
$E=10 \times 2 \times .1 $
$E=2 $
$E=2 V m$
Hence, the correct option is (B)
Question 3
1 / -0

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

A
$\frac{R}{4}$

B
$\frac{R}{2}$

C
R

D
2R

Solution

In balanced Wheatstone bridge, the galvanometer arm can be neglected so, equivalent resistance = R.

Hence, the correct option is (C)
Question 4
1 / -0

The time period of a particle in simple harmonic motion is 8 seconds. At t=0, it is at the mean position. The ratio of the distances traveled by it in the first and second seconds is:

A
1/2

B
1/√2

C
1/(√2−1)

D
1/√3

Solution

$T=8 s e c ; w=\frac{\pi}{4} r a d / s e c$
$x=A \sin w t$
At $t=1$
$x_{1}=A \sin \left(\frac{\pi}{4}\right)$
$=\frac{A}{\sqrt{2}}$
At $t=2$
$x_{2}=A \sin \left(\frac{\pi}{2}\right)=A$
$x_{2}-x_{1}=A-\frac{A}{\sqrt{2}}=A\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
$\frac{x_{1}}{x_{2}-x_{1}}=\frac{1}{\sqrt{2}-1}$
Hence, the correct option is (C)
Question 5
1 / -0

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

A
π/2

B
2π

C
π

D
π/4

Solution

In SHM $a=-\omega^{2} x$
or $a=\frac{-K}{m} x$
so, from graph $-\frac{K}{m}=-1$
$(\because$slope is-1$)$
$\frac{K^{m}}{m}=1$
Time period $=2 \pi \sqrt{\frac{m}{K}}$
$=2 \pi \sqrt{\frac{1}{1}}$
$=2 \pi$
Hence, the correct option is (B)
Question 6
1 / -0

Two simple pendulums of lengths 100 cm and 196 cm are in phase at the mean position at a certain time. If T is the time period of shorter pendulum, the minimum time after which they will be again in phase

A
2.5T

B
3.5T

C
5T

D
7T

Solution

$T=2 \pi \sqrt{\frac{l}{g}}$
$T_{2}=2 \pi \sqrt{\frac{l_{1}}{g}}$
$T_{1}=2 \pi \sqrt{\frac{100}{g}}$
$T_{2}=2 \pi \sqrt{\frac{l}{g}}$
$T_{2}=2 \pi \sqrt{\frac{196}{g}}$
$\frac{T_{1}}{T_{2}}=\frac{5}{7}$
$\therefore\left(\omega_{1}-\omega_{2}\right) t=2 \pi$
$\left(\frac{2 \pi}{T_{1}}-\frac{2 \pi}{T_{2}}\right) t=2 \pi$
$t=\frac{7}{2}$
$t=3.5 \mathrm{sec}$
Hence, the correct option is (B)
Question 7
1 / -0

A pendulum bob is in SHM. The velocity of the bob in the mean position is V. If now the amplitude of oscillation is doubled and the length of the pendulum is also doubled, the speed of the bob in the mean position would be

A
V

B
√2 V

C
V/2

D
4V

Solution

Velocity in S.H.M is: $v=A \omega$ As length is doubled $T \propto \sqrt{l}$ $\Rightarrow T_{1}=\sqrt{2} T$
$\Rightarrow \omega=\frac{2 \pi}{T}$
$\therefore v_{1}=(2 A) \frac{\omega}{\sqrt{2}}=\sqrt{2} A \omega=\sqrt{2} V$
Hence, the correct option is (B)
Question 8
1 / -0

Two particles are in S.H.M. along parallel straight lines with same amplitude and time period. If they cross each other in opposite directions at the midpoint of mean and extreme positions. Phase difference between them is:

A
30°

B
120°

C
150°

D
180°

Solution

since $x=A \sin (\omega t+\phi)$
$\frac{A}{2}=A \sin (\omega t+\phi)$
$\Rightarrow \sin (\omega t+\phi)=\frac{1}{2}$
$\Rightarrow \omega t+\phi=30^{\circ}$ or $\quad 150^{\circ}$
So the one particle has phase of $30^{\circ}$ and another has $150^{\circ}$ So the phase difference is $150^{\circ}-30^{\circ}=120^{\circ}$
Hence, the correct option is (B)
Question 9
1 / -0

The potential energy of a particle in SHM, 0.2 sec after passing the mean position is 1/4 of its total energy. The period of oscillation is:

A
3 s

B
2.4 s

C
1.8 s

D
1.2 s

Solution

$T E=\frac{1}{2} m \omega^{2} A^{2}$
$P E=\frac{1}{2} m \omega^{2} x^{2}$
$P E=\frac{T E}{4}$
$\frac{1}{2} m \omega^{2} x^{2}=\frac{1}{4} \times \frac{1}{2} m \omega^{2} A^{2}$
$x=\frac{A}{2}$
$x=A \sin \theta$
Asin\theta$=\frac{A}{2}$
$\theta=\omega t=\frac{\pi}{6}$
or $\omega=\frac{\pi}{1.2}$
Thus we get $T=\frac{2 \pi}{\omega}=\frac{2 \pi \times 1.2}{\pi}=2.4 \mathrm{sec}$
Hence, the correct option is (B)
Question 10
1 / -0

The equation of motion of a particle in S.H.M. is given by a=−Bx, where a is the acceleration, B is a constant and x is displacement. The period of oscillation of the particle is:

A
2π√B

B
2π/B

C
2π/√B

D
2 πB

Solution

$a=-B x$
so, $B=\omega^{2}\left(\therefore a=-\omega^{2} x\right)$
$T=\frac{2 \pi}{\omega}$
$=\frac{2 \pi}{\sqrt{B}}$
so,time period of oscillations$=\frac{2 \pi}{\sqrt{B}}$
Hence, the correct option is (C)

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Physics Test - 28

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Physics Test - 28

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SHARING IS CARING

Question 1

3.6 ampere

2.8 ampere

2.5 ampere

5.0 ampere

Solution

Question 2

1

2

3

4

Solution

Question 3

R4

R2

R

2R

Solution

Question 4

1/2

1/√2

1/(√2−1)

1/√3

Solution

Question 5

π/2

2π

π

π/4

Solution

Question 6

2.5T

3.5T

5T

7T

Solution

Question 7

V

√2 V

V/2

4V

Solution

Question 8

30°

120°

150°

180°

Solution

Question 9

3 s

2.4 s

1.8 s

1.2 s

Solution

Question 10

2π√B

2π/B

2π/√B

2 πB

Solution

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