Physics Test

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Physics Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Due to some force F₁ a body oscillates with period (4/5) s and the due to other force F₂ it oscillates with period 3/5s. if both the forces acts simultaneously new period will be

A
0.72s

B
0.64s

C
0.48s

D
0.36s

Solution

$F=\frac{4 \pi^{2} m x}{T^{2}}$
$F_{1}=\frac{4 \pi^{2} m x}{T^{2}}$
$F_{2}=\frac{4 \pi^{2} m x}{T_{2}^{2}}$
$F_{1}+F_{2}=\frac{4 \pi m x}{T^{2}}$
$\therefore \frac{1}{T^{2}}=\frac{1}{T_{1}^{2}}+\frac{1}{T_{2}^{2}}$
$\frac{1}{T^{2}}=\left(\frac{5}{4}\right)^{2}+\left(\frac{5}{3}\right)^{2}$
$T=0.48$
Hence, the correct option is (C)
Question 2
1 / -0

A block of mass 100 g attached to a spring of stiffness 100 N/m is lying on a frictionless floor as shown. The block is moved to compress the spring by 10 cm and released. If the collision with the wall is elastic the time period of motion is:

A
0.2 s

B
0.166 s

C
0.155 s

D
0.133 s

Solution

$5 \mathrm{cm}$ is $\frac{A}{2}$ and $O$ to $\frac{A}{2},$ the time taken is $\frac{T}{12}$ $T=t_{L H S}+t_{R H S}=\frac{T}{2}+2\left(\frac{T}{12}\right)$
$=\frac{2}{3} T=\frac{2}{3}(2 \pi) \sqrt{\frac{m}{k}}$
$=\frac{4 \pi}{3} \sqrt{\frac{0.1}{100}}=0.133 \mathrm{s}$
Hence, the correct option is (D)
Question 3
1 / -0

A body of mass 'MM' when hung on a spiral spring, the spring stretches by 20 cm, when it is pulled down and released its period of oscillation is

A
2π/5

B
2π/7

C
2π/3

D
2π

Solution

$M g=k(20) \times 10^{-2}$
$k=5 m g$
$T=2 \pi \sqrt{\frac{M}{k}}$
$T=2 \pi \sqrt{\frac{M}{5 \times M g}}$
$T=2 \pi \sqrt{\frac{1}{5 g}}$
$T=\frac{2 \pi}{7}$
Hence, the correct option is (B)
Question 4
1 / -0

A particle follows two SHM's. One along the y axis and the other along the x axis.What is the trajectory of the particle of combined SHM's. If A₁≠A₂ and ϕ= π/2 the nature of path followed by the particle is

A
straight line with positive slope

B
parabola symmetric about y-axis

C
Circle in x-y plane

D
An ellipse in x-y plane

Solution

Let equation of the particle moving along $y$ axis $\quad y=A_{1} \sin (w t)$
$\Longrightarrow$ sinwt $=\frac{y}{A_{1}}$
Now equation of the particle moving along $\mathrm{x}$ axis $\quad x=A_{2} \sin (w t+\phi)$
Given : $\phi=\frac{\pi}{2}$
Thus $\quad x=A_{2} \sin \left(w t+\frac{\pi}{2}\right)=A_{2} \cos (w t)$
$\ldots \ldots$
$\Longrightarrow \cos w t=\frac{x}{A_{2}}$
Squaring and adding
(1) and (2) $\sin ^{2} w t+\cos ^{2} w t=\frac{y^{2}}{A_{1}^{2}}+\frac{x^{2}}{A_{2}^{2}}$
$\therefore \quad \frac{y^{2}}{A_{1}^{2}}+\frac{x^{2}}{A_{2}^{2}}=1 \quad$ which is an equation
of an ellipse.
Hence, the correct option is (D)
Question 5
1 / -0

If a particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P (AP

A
3 m/s

B
$3 \sqrt{2}$ m/s

C
$3 \sqrt{3}$ m/s

D
6 m/s

Solution

$V=\omega \sqrt{A^{2}-x^{2}}$
$\omega \times t=\theta$
so,
$\omega \times 0.5=\theta$
$\omega \times 1.5=(2 \pi-\theta)$
on adding the above equations; $\omega \times 2=2 \pi$
$\omega=\pi$
$\therefore \theta=\frac{\pi}{4}$
$x=A \cos \theta$
$\frac{3=\pi \sqrt{A^{2}-\frac{A^{2}}{2}}}{V=\pi A}$
$\therefore V_{\max }=3 \sqrt{2} m / \sec$
Hence, the correct option is (B)
Question 6
1 / -0

It is cut into three parts with lengths in the ratio 1:2:3 and connected to the mass in a particular combination so that its time period is halved, what is that combination?

A
First two parts in series and the third one should be parallel to first two.

B
First two should be parallel and this combination in series with third one.

C
All should be in parallel.

D
All should be in series.

Solution

$K=\frac{E A}{l}$ Let $\mathrm{K}$ be the spring constant of original spring $1^{\text {st}}$ part $K_{1}=\frac{E A}{\frac{l}{6}}=6 \frac{E A}{l}=6 K$
$2^{\text {nd }}$ part $K_{2}=\frac{E A}{\frac{2 l}{6}}=3 \frac{E A}{l}=3 K$
$3^{\text {rd}}$ part
$K_{3}=\frac{E A}{\frac{3 l}{6}}=2 \frac{E A}{l}=2 K$
Time period $T_{e f f}=2 \pi \sqrt{\frac{m}{K_{e f f}}}$
If $K_{e f f}=4 K$ then $T_{e f f}=\frac{T}{2}$
For $K_{e f f}$ to be $4 \mathrm{K}$ $K_{1}$ and $K_{2}$ should be in series Let effective of $K_{1}$ and $K_{2}=K_{4}$ $\frac{1}{K_{4}}=\frac{1}{K_{1}}+\frac{1}{K_{2}}=\frac{1}{6 K}+\frac{1}{3 K} ; K_{4}=2_{K}$
Effective of $K_{4}$ and $K_{3}=K_{4}+K_{3}=2 K+2 K=4 K$
Hence, the correct option is (A)
Question 7
1 / -0

The time periods of oscillation of two simple pendulums are 1 sec, 1.2 sec. Initially both are in same phase of oscillation. The minimum number of oscillations made by longer pendulum when they are again in same phase is:

A
5

B
6

C
10

D
12

Solution

$t\left(\omega_{1}-\omega_{2}\right)=2 \pi$
$t=\frac{T_{1} T_{2}}{T_{2}-T_{1}}$
$=\frac{1 \times 1.2}{1.2-1}$
$t=6$
$\therefore$ no. of oscillations made lag longer pendulum $=\frac{6}{1.2}$
$=5$
Hence, the correct option is (A)
Question 8
1 / -0

Force acting on a block is F =(−4x+8)=. Here F is in Newton and xx the position of block on x-axis in meters_________

A
motion of the block is periodic but not simple harmonic

B
motion of the block is not periodic

C
motion of the block is simple harmonic about origin, x=0

D
motion of the block is simple harmonic about x=2m.

Solution

Block has its SHM mean about the points where net force acting on it is Zero.

If is Zero about x=2m

Thus SHM has mean position at x=2m
Hence, the correct option is (D)
Question 9
1 / -0

The change in the length of a simple pendulum of length 1m, when its period of oscillation changes from 2s to 1.5s is:

A
increased by (7/8)m

B
decreased by $4 \frac{7}{8} m$

C
increased by (7/16)m

D
decreased by (7/16)m

Solution

$T_{1}=2 \sec$
$L_{1}=1 m$
$T_{2}=1.5 \mathrm{sec}$
$T=2 \pi \sqrt{\frac{l}{g}}$
$T \alpha \sqrt{L}$
$\frac{T_{1}}{\sqrt{L_{2}}}=\frac{T_{2}}{\sqrt{L_{2}}}$
$\therefore \frac{2}{\sqrt{1}}=\frac{1.5}{\sqrt{L_{2}}}$
$L_{2}=\frac{9}{16}$
$L_{1}-L_{2}=1-\frac{9}{16}$
$=\frac{7}{16}$
$\therefore$ length decreased by $\frac{7}{16} m$
Hence, the correct option is (D)
Question 10
1 / -0

The displacement of a particle of mass 3 gm executing simple harmonic motion is given by y=3sin(0.2t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to 1/3 of its amplitude from its mean position is

A
12 x 10⁻³ J

B
25 x 10⁻³J

C
0.480 x 10⁻³J

D
0.24 x 10⁻³ J

Solution

$y=$Asinwt
$y=3 \sin (0.2 t)$
$A=3 m$
$\omega=0.2 \mathrm{rad} / \mathrm{sec}$
$m=3 \times 10^{-3} \mathrm{kg}$
$K E=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$
$=\frac{1}{2} 3 \times 10^{-3}(0.2)^{2}\left(3^{2}-\left(3 \times \frac{1}{3}\right)^{2}\right)$
$=\frac{1}{2} \times 3 \times 10^{-3} \times(0.2)^{2} \times 8$
$=4.8 \times 10^{-3}$
Hence, the correct option is (C)

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Physics Test - 30

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Physics Test - 30

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SHARING IS CARING

Question 1

0.72s

0.64s

0.48s

0.36s

Solution

Question 2

0.2 s

0.166 s

0.155 s

0.133 s

Solution

Question 3

2π/5

2π/7

2π/3

2π

Solution

Question 4

straight line with positive slope

parabola symmetric about y-axis

Circle in x-y plane

An ellipse in x-y plane

Solution

Question 5

3 m/s

32m/s

33m/s

6 m/s

Solution

Question 6

First two parts in series and the third one should be parallel to first two.

First two should be parallel and this combination in series with third one.

All should be in parallel.

All should be in series.

Solution

Question 7

5

6

10

12

Solution

Question 8

motion of the block is periodic but not simple harmonic

motion of the block is not periodic

motion of the block is simple harmonic about origin, x=0

motion of the block is simple harmonic about x=2m.

Solution

Question 9

increased by (7/8)m

decreased by \(4 \frac{7}{8} m\)

increased by (7/16)m

decreased by (7/16)m

Solution

Question 10

12 x 10−3 J

25 x 10−3 J

0.480 x 10−3 J

0.24 x 10−3 J

Solution

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$3 \sqrt{2}$ m/s

$3 \sqrt{3}$ m/s

12 x 10⁻³ J

25 x 10⁻³J

0.480 x 10⁻³J

0.24 x 10⁻³ J