Physics Test

Result

Physics Test - 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

The excess pressure inside a spherical drop of water is four times that of another drop. Then, their mass ratio is

A
1 : 16

B
8 : 1

C
1 : 4

D
1 : 64

Solution

\(\frac{2 T}{r_{1}}=4 \times \frac{2 T}{r_{2}} \Rightarrow r_{2}=4 r_{1}\)
Mass ratio \(=M_{1}: M_{2}=r_{1}^{3}: r_{2}^{3}=1: 64\)
Question 2
4 / -1

A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s, a stone is released from the balloon. The stone will, after release from the balloon,

A
cover a distance of 40 m

B
have a displacement of 50 m

C
reach the ground in 4 s

D
begin to move down

Solution

Velocity of the balloon after 8 sec
\(\mathrm{v}=\mathrm{u}+\mathrm{at}=0+1.25 \mathrm{x} 8=10 \mathrm{m} / \mathrm{s}\)
Height attained by the balloon after 8 sec \(\mathrm{H}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=0+\frac{1}{2} \times 1.25 \times\left(8 \mathrm{s}^{2}=40 \mathrm{m}\right.\)
Stone will have same velocity as that of balloon, after 8 sec, the stone will travel under acceleration due
to gravity.
Distance travelled by the stone before coming to rest \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as} \Rightarrow \mathrm{s}=\frac{10 \times 10}{2 \times 10}=5 \mathrm{m}\)
Time taken by the stone to reach the ground
Displacement \(=-40 \mathrm{m}, \mathrm{u}=+10 \mathrm{m} / \mathrm{s}\) and \(\mathrm{a}=-10 \mathrm{m} / \mathrm{s}^{2}\)
\(-40=10 t-\frac{1}{2} \times 10 \times t^{2} \Rightarrow-8=2 t-t^{2}\)
\(t^{2}-2 t-8=0\)
\(\Rightarrow(t-4)(t+2)\)
Neglecting the negative value, we get \(t=4\) sec
Question 3
4 / -1

A particle in SHM is described by the displacement equation x(t) = A cos (ωt + θ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm/s, then what is its amplitude? (The angular frequency of the particle is πs^-1.)

A
1 cm

B
√2 cm

C
2 cm

D
2.5 cm

Solution

Rate of change of displacement gives velocity. Given, \(\quad x=A \cos (\omega t+\theta)\)
Velocity \(v=\frac{d x}{d t}=A \frac{d}{d t} \cos (\omega t+\theta)\)
\(v=-A \omega \sin (\omega t+\theta)\)
Using \(\sin ^{2} \theta+\cos ^{2} \theta=1,\) we have
\(v=-A \omega \sqrt{1-\cos ^{2}(\omega t+\theta)}\)
\(\therefore\) From equation \(x=A \cos (\omega t+\theta),\) we have \(v=-A \omega \sqrt{1-x^{2} / A^{2}}\)
\(\Rightarrow \quad v=-\omega \sqrt{A^{2}-x^{2}}\)
Given, \(v=\pi \mathrm{cm} / \mathrm{s}, x=1 \mathrm{cm}, \omega=\pi \mathrm{s}^{-1}\)
\(\therefore \quad \pi=-\pi \sqrt{A^{2}-1}\)
\(\Rightarrow \quad 1=A^{2}-1\)
\(\Rightarrow A=\sqrt{2} \mathrm{cm}\)
Question 4
4 / -1

From the top of a tower, a stone is thrown up and reaches the ground in time t₁ = 9 s. A second stone is thrown down with the same speed and reaches the ground in time t₂ = 4 s. A third stone is released from rest and reaches the ground in time t₃, which is equal to

A
6.5 s

B
6 s

C
65 s

D
64 s

Solution

Taking the downward direction as positive Displacement of first stone,
\[
h=-u t_{1}+\frac{1}{2} g t_{1}^{2}
\]
Displacement of second stone,
\[
h=u t_{2}+\frac{1}{2} g t_{2}^{2}
\]
Displacement of third stone,
\[
h=\frac{1}{2} g t_{3}^{2}
\]
Multiplying Eq. (1) by \(t_{2}\) and Eq. (2) by \(t_{1}\) and adding, we get
\[
h\left(t_{1}+t_{2}\right)=\frac{1}{2} g t_{1} t_{2}\left(t_{1}+t_{2}\right)
\]
\(\Rightarrow \quad h=\frac{1}{2} g t_{1} t_{2}\)
From Eqs. (3) and (4), \(t_{3}^{2}=t_{1} t_{2}\)
or \(t_{3}=\sqrt{t_{1} t_{2}}=\sqrt{9 \times 4}=6 \mathrm{s}\)
Question 5
4 / -1

Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in the figure. Ball 1 moves with a velocity v₁toward balls 2 and 3. All collisions are assumed to be elastic. If M < m, the number of collisions between the balls will be

A
one

B
two

C
three

D
four

Solution

The first collision will be between balls 1 and \(2 .\) since both have the same mass, after the collision ball 1
win come to rest and ball 2 win move with speed \(v_{1} .\) This ball will collide with the stationary ball 3 .
Alter this second collision, let \(v_{2}\) and \(v_{3}\) be the speeds of balls 2 and 3 respectively. since the collisions
are elastic, \(\mathrm{v}_{2}\) and \(\mathrm{v}_{3}\) are given by
\[
v_{2}=\left(\frac{m-M}{m+M}\right) v_{1}
\]
(i)
and \(v_{3}=\left(\frac{2 m}{m+M}\right) v_{1}\)
(ii)
If \(\mathrm{M}<\mathrm{m},\) it follows from (i) and (ii) that \(\mathrm{v}_{2}<\mathrm{v}_{3}\) and both have the same direction. Therefore, ball 2
cannot collide with ball 3 again. Hence there are only two collisions.
Question 6
4 / -1

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. The force constant of the longer piece will be

A
1.5k

B
3k

C
2k

D
2/3k

Solution

The force constant of a spring is inversely propor. tional to its length. If a spring of length \(L\) is cut into two pieces of lengths \(x\) and \((L-x)\), such that
\[
x=2(L-x) \text { or } x=\frac{2 L}{3}
\]
then the force constant of the spring of length \(x\) is related to the force constant \(k\) of the complete spring of length \(L\) as
\[
\frac{k_{1}}{k}=\frac{L}{x}=\frac{L}{2 L / 3}=\frac{3}{2} \text { or } k_{1}=\frac{3}{2} k
\]
Question 7
4 / -1

A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s2 due west. The distance covered by the particle in the fifth second of its motion is

A
0 m

B
0.5 m

C
2 m

D
None of these

Solution

The displacement is indeed 0 but not the distance. the distance covered is 0.5 m.

Actually, the car is changing the direction of motion at 4.5 seconds. As such displacement between 4 and 5 seconds is 0 but still the car covers some distance.

The distance is the mod of displacement made by the car between 4 and 4.5 seconds + mod of the displacement between 4.5 and 5 seconds.

Displacement b/w 4 to 4.5 seconds = (9)(0.5) - (4.5² - 4²) = 4.5 - 4.25 = 0.25

Displacement b/w 4.5 to 4 seconds = (9)(0.5) - (5² - 4.5²) = 4.5 - 4.75 = -0.25

Displacement = 0, Distance = 0.5 m
Question 8
4 / -1

A block released on a rough inclined plane of inclination = 30° slides down the plane with an acceleration g/4, where g is the acceleration due to gravity. What is the coefficient of friction between the block and the inclined plane?

A
2/√3

B
1/√3

C
1/2√3

D
√3/2

Solution

Acceleration of an object down the incline is 'a
ma \(=m g \sin \theta-f\)
\(f=\mu N=\mu m g \cos \theta\)
\(a=g \sin \theta-\mu g \cos \theta\)
\(\theta=30^{\circ}\)
\(\frac{9}{4}=g\left(\sin 30^{\circ}-\mu \cos 30^{\circ}\right)\)
\(\frac{1}{2}-\mu \frac{\sqrt{3}}{2}=\frac{1}{4} \Rightarrow \mu \frac{\sqrt{3}}{2}=\frac{1}{4}\)
\(\mu=\frac{1}{2 \sqrt{3}}\)
Question 9
4 / -1

A boy, while catching a ball, experiences an impulse of 6 Ns. If the mass of the ball is 200 g, then what was the speed of the ball before it was caught?

A
10 ms^-1

B
30 ms^-1

C
20 ms^-1

D
40 ms^-1

Solution

Impulse = Change in momentum
Let v be the speed of the ball before it was caught and \(m\) be its mass.
since the ball is brought to rest after it is caught, the change in momentum \(=\mathrm{mv}\). Now, \(m=0.2 \mathrm{kg} .\) Therefore, impulse is \(0.2 \mathrm{v}\). Given, impulse \(=6 \mathrm{Ns}\) Hence, \(v=\frac{6}{0.2}=30 \mathrm{ms}^{-1}\)
Thus, the correct choice is (2) .
Question 10
4 / -1

The angular velocity of a planet revolving in an elliptical orbit around the sun __________ when it comes near the sun

A
decreases

B
increases

C
remains the same

D
None of these

Solution

If we consider planet and the sun as one system, then the force of gravitation between the sun and the
planet is a central force hence no torque is generated about sun ..It means angular momentum of the planet is consereved \(L=I \omega=\) costant. As the planet comes near to the sun, Moment of I nertia
decreases which means there will be increase in the angular veloicty of the planet.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 22

Result

Physics Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

1 : 16

8 : 1

1 : 4

1 : 64

Solution

Question 2

cover a distance of 40 m

have a displacement of 50 m

reach the ground in 4 s

begin to move down

Solution

Question 3

1 cm

√2 cm

2 cm

2.5 cm

Solution

Question 4

6.5 s

6 s

65 s

64 s

Solution

Question 5

one

two

three

four

Solution

Question 6

1.5k

3k

2k

2/3k

Solution

Question 7

0 m

0.5 m

2 m

None of these

Solution

Question 8

2/√3

1/√3

1/2√3

√3/2

Solution

Question 9

10 ms-1

30 ms-1

20 ms-1

40 ms-1

Solution

Question 10

decreases

increases

remains the same

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

10 ms^-1

30 ms^-1

20 ms^-1

40 ms^-1