Physics Test

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Physics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

A simple pendulum whose length is slightly less than that of a seconds pendulum start vibrating with another seconds pendulum in the same phase. After 22 seconds, they again vibrate in the same phase. The periodic time of simple pendulum is-

A
1.09 sec

B
1.4 sec

C
1.6 sec

D
1.9 sec

Solution

It is mentioned that, simple pendulum whose length is slightly less than that of seconds pendulum starts vibrating with another seconds pendulum after 18 seconds they again vibrate in the same phase. Let $T = 22 s$ , the time period of the simple pendulum is $2 s$ , $T_{2} = 2 s$ .

It is required to find the periodic time. Let it is given by $T_{1}$ . So,

$\begin{array}{l} \frac{1}{T} = \frac{1}{T_{1}} - \frac{1}{T_{2}} \\ \frac{1}{T_{1}} = \frac{1}{T} + \frac{1}{T_{2}} \end{array}$

$\frac{1}{T_{1}} = \frac{1}{22} + \frac{1}{2}$

$T_{1} = 1.833 s$

So, the periodic time is$1.833 \sim 1.9 \mathrm{s}$
Question 2
1 / -0

If the radius of Earth decreases by $1\%$ and its mass remains the same, then the acceleration due to gravity:

A
increases by $1\%$

B
decreases by $1\%$

C
increases by $2\%$

D
decreases by $2\%$

Solution

Given: $\frac{\Delta R}{R} \times 100=-1 \%$; $M$ remains same.

As we know, $g=\frac{G M}{R^{2}}$

Taking $\log$ and differentiating,

$\frac{\Delta g}{g}=-2 \frac{\Delta R}{R}$

$\frac{\Delta g}{g} \times 100=-2\left(\frac{\Delta R}{R} \times 100\right)$

$=-2(-1)\%$

$=+2\%$

Positive sign indicates the increase in the value of $g$.
Question 3
1 / -0

If the radius of Earth suddenly decreases to $80 \%$ of its present value (the mass of Earth remaining the same), then the value of acceleration due to gravity will:

A
remain unchanged

B
become $(9.8 \times 0.8) ms ^{-2}$

C
increase by $36 \%$

D
increase by about $56 \%$

Solution

Now, $g=\frac{G M}{R^{2}}$.

If $R$ reduces to $R=0.8 R$, the value of $g$ becomes

$g' =\frac{G M}{R^{2}}$

$=\frac{G M}{0.64 R^{2}}$

$=\frac{g}{0.64}$

$=\frac{9.8 m s^{-2}}{0.64}$

Increase in value of $g =g'-\frac{1}{g}$

$=\frac{0.36 g }{0.64}$

Percentage increase

$=\frac{0.36 g }{0.64 g } \times 100$

$=56.25 \%$
Question 4
1 / -0

If a person goes to a depth equal to the radius of Earth below its surface, then what would be his weight relative to that on Earth?

A
$\frac{1}{4}$

B
$\frac{1}{8}$

C
$0$

D
$\frac{1}{16}$

Solution

Let $g ^{\prime}$ be the gravitational acceleration with variation in depth

$g^{\prime}=\left(1-\frac{h}{R}\right) g$

$w^{\prime}=m g^{\prime}=m\left(1-\frac{h}{R}\right) g$

At depth,

$h=R$

$\Rightarrow w^{\prime}=0$

Therefore, it is zero at equally below the surface.
Question 5
1 / -0

In Nehru Planetarium, an astronomical telescope which has an objective lens of diameter 10 m. To see anything through this telescope, planetarium team glow a light of wavelength 7000Α°. What will be the resolving power of this telescope?

A
$54.6 \times 10^{6}$

B
$23.7 \times 10^{5}$

C
$11.7 \times 10^{6}$

D
$42.6 \times 10^{5}$

Solution

Resolving power (RP) of telescope is

$R P = \frac{D}{1.22 λ}$

$R P = \frac{10}{1.22 \times 7000 \times 10^{- 10}}$

$R P = 11.7 \times 10^{6}$

Hence option (C) is correct.
Question 6
1 / -0

For a lens of simple microscope, with magnifying power 2, what is the minimum value of distance for clear vision if the focal length of the lens is 20 m?

A
10m

B
20m

C
30m

D
40m

Solution

We know,

If magnifying power is MP, distance of minimum vision is $D$ , and focal length is $f$ , then,

$M P = (1 + \frac{D}{f})$

Therefore,

$D = (M P - 1) f$

Thus,

$D = (2 - 1) 20 = 20 m$

$D = 20 m$

Hence option (B) is correct
Question 7
1 / -0

A disc has a speed of $1200 rpm$ and it is made to slow down at a uniform rate of $4 rad/s^2$. Calculate the number of revolution it makes before coming to rest?

A
$314$

B
$215$

C
$328$

D
$285$

Solution

$\mathrm{f}=1200 \mathrm{rpm}=20$ rotation per sec Initial angular velocity of the disc $\left(\omega_{0}\right)=2 \pi f=2 \pi \times 20=40 \pi \mathrm{rad} / \mathrm{s}$
Final angular velocity of the disc $\left(\omega_{f}\right)=0$ Angular acceleration $\left(\alpha)=-4 \mathrm{rad} / \mathrm{s}^{2}\right.$
Now
$\omega_{f}=\omega_{0}+\alpha t$
or $t=\frac{\omega_{0}}{\alpha}=10 \pi \mathrm{sec}$
Now $\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$
$\theta=200 \pi^{2}$ rad
Therefore, Number of revolutions $=\frac{200 \pi^{2}}{2 \pi}=314$

Hence the correct answer is option (a).
Question 8
1 / -0

A body is projected at time $t=0$ from a certain point on a planet's surface with a certain velocity at a certain angle with the planet's surface (assumed horizontal). The horizontal and vertical displacement $x$ and $y$ (in metre ) respectively vary with time $t$ in second as, $x=(10 \sqrt{3}) t$ and $y=10 t-t^{2}$. The maximum height attained by the body is:

A
$25 m$

B
$50 m$

C
$75 m$

D
$100 m$

Solution

For maximum value of $y$

$\frac{d y}{d t}=10-2 t=0$

$\Rightarrow t=5 s$

$y_{\max }=(10)(5)-(5)^{2}$

$=25 m$
Question 9
1 / -0

Two balls are projected simultaneously in the same vertical plane from the same point with velocities $V _{1}$ and $V _{2}$ at angles $\theta_{1}$ and $\theta_{2}$, respectively, with the horizontal. If $V _{1} \cos \theta_{1}= V _{2} \cos \theta_{2}$, the path of one ball as seen from the position of other ball is:

A
parabola

B
horizontal straight line

C
vertical straight line

D
straight line making $45^{\circ}$ with the vertical

Solution

For the ball projected with velocity $V _{1}$ at an angle $\theta_{1}$ with horizontal line, the horizontal distance covered after $t$ time.

$x_{1}=V_{1} \cos \theta_{1} t$

Similarly, for second ball throw with velocity $V _{2}$ at an angle $\theta_{2}$ with horizontal, horizontal distance covered after time $t .$

$x_{2}=V_{2} \cos \theta_{2} t$

The vertical distance covered are

$y_{1}=V_{1} \sin \theta_{1} t-\frac{1}{2} g t^{2}$

and $ y_{2}=V_{2} \sin \theta_{2} t-\frac{1}{2} g t^{2}$

$\therefore x_{2}-x_{1}=\left(V_{2} \cos \theta_{2}-V_{1} \cos \theta_{1}\right) t$

and $y_{2}-y_{1}=\left(V_{2} \sin \theta_{2}-V_{1} \sin \theta_{1}\right) t$

$\therefore \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{V_{2} \sin \theta_{2}-V_{1} \sin \theta_{1}}{V_{2} \cos \theta_{2}-V_{1} \cos \theta_{1}}$

but $V_{1} \cos \theta_{1}=V_{2} \cos \theta_{2}$

$\therefore \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{V_{2} \sin \theta_{2}-V_{1} \sin \theta_{1}}{0}=\infty$

$\Rightarrow x_{2}-x_{1}=0$

and $ y_{2}-y_{1}=\infty$

This means line joining the position of particles after time $t$ will be a straight line and parallel to the $y$-axis.
Question 10
1 / -0

A body is projected at time $t=0$ from a certain point on a planet's surface with a certain velocity at a certain angle with the planet's surface (assumed horizontal). The horizontal and vertical displacements $x$ and $y$ (in metres) respectively vary with time $t$ (in seconds) as: $x=10 \sqrt{3} t$ and $y=10 t-t^{2}$

What is the magnitude and direction of the velocity with which the body is projected?

A
$20 ms ^{-1}$ at an angle of $30^{\circ}$ with the horizontal

B
$20 ms ^{-1}$ at an angle of $60^{\circ}$ with the horizontal

C
$10 ms ^{-1}$ at an angle of $30^{\circ}$ with the horizontal

D
$10 ms ^{-1}$ at an angle of $60^{\circ}$ with the horizontal

Solution

We know that the position coordinates $x$ and $y$ are given by

$x=(u \cos \theta) t$.............(i)

and $y=(u \sin \theta) t-\frac{1}{2} g t^{2}$................(ii)

Comparing equation (1) with $x=10 \sqrt{3} t$ we have,

$u \cos \theta=10 \sqrt{3} \mathrm{ms}^{-1}$

Also, comparingequation (ii) with $y=10 t-t^{2}$ we have,

$u \sin \theta=10 {ms}^{-1}$

These equations give $u^{2}=10^{2}+10^{2} \times 3=400$

or $u=20 {ms}^{-1}$

and $\tan \theta=\frac{1}{\sqrt{3}}$ which gives $\theta=30^{\circ}$

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Physics Test - 6

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Question 1

1.09 sec

1.4 sec

1.6 sec

1.9 sec

Solution

Question 2

increases by \(1\%\)

decreases by \(1\%\)

increases by \(2\%\)

decreases by \(2\%\)

Solution

Question 3

remain unchanged

become \((9.8 \times 0.8) ms ^{-2}\)

increase by \(36 \%\)

increase by about \(56 \%\)

Solution

Question 4

\(\frac{1}{4}\)

\(\frac{1}{8}\)

\(0\)

\(\frac{1}{16}\)

Solution

Question 5

54.6×106

23.7×105

11.7×106

42.6×105

Solution

Question 6

10m

20m

30m

40m

Solution

Question 7

\(314\)

\(215\)

\(328\)

\(285\)

Solution

Question 8

\(25 m\)

\(50 m\)

\(75 m\)

\(100 m\)

Solution

Question 9

parabola

horizontal straight line

vertical straight line

straight line making \(45^{\circ}\) with the vertical

Solution

Question 10

\(20 ms ^{-1}\) at an angle of \(30^{\circ}\) with the horizontal

\(20 ms ^{-1}\) at an angle of \(60^{\circ}\) with the horizontal

\(10 ms ^{-1}\) at an angle of \(30^{\circ}\) with the horizontal

\(10 ms ^{-1}\) at an angle of \(60^{\circ}\) with the horizontal

Solution

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$54.6 \times 10^{6}$

$23.7 \times 10^{5}$

$11.7 \times 10^{6}$

$42.6 \times 10^{5}$