Physics Test

Result

Physics Test - 11

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
1 / -0

Two blocks of masses m₁= m and m₂ = 2m are connected by a light string passing over a frictionless pulley. The mass m1 is placed on a smooth inclined plane of inclination $θ$ = 30^o and mass m₂ hangs vertically as shown in the figure.

If the system is released, the blocks move with an acceleration equal to:

A
g/4

B
g/3

C
g/2

D
g

Solution

the equation of the block are
$T-m_{1} g \sin \theta=m_{1} a \ldots \ldots$(1)
$m_{2} g-T=m_{2} a \ldots \ldots \ldots$(2)
Equation (1) and (2)
$a=\frac{\left(m_{2}-m_{1} \sin \theta\right) g}{\left(m_{1}+m_{2}\right)}$
$=\frac{\left(2 m-m \times \sin 30^{\circ}\right) g}{(m+2 m)}$
$=\frac{g}{2}$
Hence the correct option is $C$
Question 2
1 / -0

Water is falling on the blades of a turbine at the rate of 6000 kg/min. The height of the fall is 100 m. The power given to the turbine is (Take g = 10 m/s²)

A
10 kW

B
6 kW

C
100 kW

D
150 kW

Solution

Mass of water falling on the turbine, $m=6 \times 10^{3} \mathrm{kg}$
Rate at which the water is flowing, $t=1 \min =60$ s
Height of the fall, $h=100 \mathrm{m}, g=10 \mathrm{m} / \mathrm{s}^{2}$
So, power given to the turbine, $P=\frac{\text { Work done }}{\text { time taken }}$
$=\frac{m g h}{t}=\frac{6 \times 10^{3} \times 10 \times 100}{60}=10^{5} W=100 \mathrm{kW}$
Question 3
1 / -0

A small ball is pushed from a height 'h' along a smooth hemispherical bowl. With what speed should the ball be pushed, so that it just reaches the top of the opposite end of the bowl? (The height of the top of the bowl is R.)

A
$\sqrt{29 h}$

B
$\sqrt{2 g R}$

C
$\sqrt{2 g(R+h)}$

D
$\sqrt{2 g(R-h)}$

Solution

when the ball $n$ at height $h$ the $P . E_{1}$ of the ball in $m g h$
when the ball is pushed with velocity $v$ it reaches
The bottom with $K . E_{1}=\frac{1}{2} m v^{2}$
Thus by inertia the ball Travel upto $R$ when its
$P . E_{2}=m g R$ and velocity $=0$
$K \cdot E_{2}=0$
$\therefore$ From low of conservation of energy
$P E_{1}+K E_{1}=P E_{2}+K E_{2}$
$m g h+\frac{1}{2} m v^{2}=m g R+0$
so $\therefore v=\sqrt{2 g(R-h)}$
Question 4
1 / -0

A body of mass m accelerates uniformly from rest to v₁ in time t₁ . As a function of t, the instantaneous power delivered to the body is:

A

$\frac{m v_{1} t}{t_{2}}$

B

$\frac{m v_{1}^{2} t}{t_{1}}$

C

$\frac{m v_{1} t^{2}}{t_{1}}$

D

$\frac{m v_{1}^{2} t}{t_{1}^{2}}$

Solution

From $v=u+a t, v_{1}=0+a t_{1}$
$\therefore a=\frac{v_{1}}{t_{1}}$
$F=m a=m v_{1} / t_{1}$
Velocity acquired in $\mathrm{t} \mathrm{sec}=a t=\frac{v_{1}}{t_{1}} t$
Power$=F \times v=\frac{m v_{1}}{t_{1}} \times \frac{v_{1} t}{t_{1}}=\frac{m v_{1}^{2} t}{t_{1}^{2}}$
Question 5
1 / -0

A ball P of mass 2 kg undergoes an elastic collision with another ball Q at rest. After the collision, ball P continues to move in its original direction with a speed one-fourth of its original speed. What is the mass of ball Q?

A
0.9 kg

B
1.2 kg

C
1.5 kg

D
1.8 kg

Solution

By conservation of linear momentum
$2 v_{0}=2\left(\frac{v_{0}}{4}\right)+m v \Rightarrow 2 v_{0}=\frac{v_{0}}{2}+m v \Rightarrow \frac{3 v_{0}}{2}=m v \ldots$
since collision is elastic
$v_{\text {separation }}=v_{\text {approach }} \Rightarrow v-\frac{v_{0}}{4}=v_{0} \Rightarrow \frac{5 v_{0}}{4}=v \ldots .(2)$
equating (2) and (1)
$\frac{3 v_{0}}{4}=m\left(\frac{5 v_{0}}{4}\right) \Rightarrow m=\frac{6}{5}=1.2 \mathrm{kg}$
Question 6
1 / -0

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ change adiabatically from $\left(\mathrm{P}_{1}, \rho_{1}\right)$ to $\left(\mathrm{P}_{2}, \rho_{2}\right)$. If $\frac{\rho_{2}}{\rho_{1}}=32$, then what should be the value of $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}$ ?

A
16

B
32

C
64

D
128

Solution

For an adiabatic process.
$P V^{\gamma}=$ constant
$P \rho^{-\gamma}=$ constant
$\frac{P_{2}}{P_{1}}=\left(\frac{\rho_{1}}{\rho_{2}}\right)^{-\gamma}=\left(\frac{1}{32}\right)^{-7 / 5}=(32)^{7 / 5}=128$
Question 7
1 / -0

Kepler's second law states that the straight line joining a planet to the sun sweeps out equal areas in equal time. This statement is equivalent to the statement that

A
total acceleration is zero

B
tangential acceleration is zero

C
longitudinal acceleration is zero

D
radial acceleration is zero

Solution

Velocity of particle accelerating uniformly in time $t_{1}$
$v=a t_{1}$
$a=\frac{v}{t_{1}}$
Velocity of particle in time $t$
$v^{\prime}=a t=\frac{v t}{t_{1}}$
According to work-energy theorem, work done = change in kinetic energy i.e., $\quad W_{\mathrm{ext}}=\Delta K$ or $W_{\text {ext }}=\frac{1}{2} m v^{2}-0$
$=\frac{1}{2} m\left(\frac{v t}{t_{1}}\right)^{2}$
$=\frac{1}{2} \frac{m v^{2}}{t_{1}^{2}} t^{2}$
Question 8
1 / -0

Two bodies A and B have masses 'M' and 'm' respectively, where M > m, and they are at a distance 'd' apart. Equal forces are applied to them so that they approach each other. The position where they hit each other is

A
near to B

B
near to A

C
equidistant from A and B

D
not determined

Solution

since the collision is elastic, both momentum and kinetic energy are conserved. If $m$ is mass of ball  and v' its speed after the collision, the law of conservation of momentum gives
$2 v=2 \times \frac{v}{4}+m v^{\prime}$
where $v$ is the original speed of ball $P$. Thus
$m v^{\prime}=\frac{3 v}{2}$ or $v^{\prime}=\frac{3 v}{2 m}$
The law of conservation of energy gives
$\frac{1}{2} \times 2 \times v^{2}=\frac{1}{2} \times 2 \times\left(\frac{v}{4}\right)^{2}+\frac{1}{2} m v^{\prime 2}$
or $m v^{2}=\frac{15 v^{2}}{8}$
Using
(i) in (ii) we get $m=1.2 \mathrm{kg}$
Question 9
1 / -0

Two satellites of Earth, S₁ and S₂, are moving in the same orbit. The mass of S₁ is four times the mass of S₂. Which of the following statements is true?

A
The time period of S₁ is four times that of S₂.

B
The potential energy of Earth and satellite in both the cases is the same.

C
S₁ and S₂ are moving with the same speed.

D
The kinetic energies of the two satellites are equal.

Solution

W = F.s = (ma).s
$W=m \frac{v}{t} s=m \frac{8}{t^{2}} s=m \frac{s^{2}}{t^{2}} \quad$ (where $m$ is the mass, $s$ is the displacement and $t$ is the time)
Now, mass, length and time are doubled. $\mathrm{SO}_{\text {, }}$
$\mathrm{W}^{\prime}=2 \mathrm{m} \frac{4 \mathrm{s}^{2}}{4 t^{2}}=2 \mathrm{m} \frac{\mathrm{s}^{2}}{\mathrm{t}^{2}}=2 \mathrm{W}$
Question 10
1 / -0

The mass of the moon is $\frac{1}{12} \mathrm{M}_{\mathrm{e}}$. If the acceleration due to gravity on the moon is $\frac{\mathrm{g}}{6},$ then its radius is

A
$\sqrt{2} \mathrm{R}_{\mathrm{e}}$R_e

B
$\frac{R_{e}}{\sqrt{2}}$

C
R_e

D
$\frac{\sqrt{2}}{R_{e}}$

Solution

The total energy $=\mathrm{KE}+\mathrm{PE}$ remains constant during the free fall, i.e.
\[
\begin{aligned}
m g h+\frac{1}{2} & m v^{2}=\mathrm{constant} \\
g h+\frac{v^{2}}{2} &=\mathrm{constant}
\end{aligned}
\]

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 11

Result

Physics Test - 11

Score

-

Rank

-

SHARING IS CARING

Question 1

g/4

g/3

g/2

g

Solution

Question 2

10 kW

6 kW

100 kW

150 kW

Solution

Question 3

\(\sqrt{29 h}\)

\(\sqrt{2 g R}\)

\(\sqrt{2 g(R+h)}\)

\(\sqrt{2 g(R-h)}\)

Solution

Question 4

\(\frac{m v_{1} t}{t_{2}}\)

\(\frac{m v_{1}^{2} t}{t_{1}}\)

\(\frac{m v_{1} t^{2}}{t_{1}}\)

\(\frac{m v_{1}^{2} t}{t_{1}^{2}}\)

Solution

Question 5

0.9 kg

1.2 kg

1.5 kg

1.8 kg

Solution

Question 6

16

32

64

128

Solution

Question 7

total acceleration is zero

tangential acceleration is zero

longitudinal acceleration is zero

radial acceleration is zero

Solution

Question 8

near to B

near to A

equidistant from A and B

not determined

Solution

Question 9

The time period of S1 is four times that of S2.

The potential energy of Earth and satellite in both the cases is the same.

S1 and S2 are moving with the same speed.

The kinetic energies of the two satellites are equal.

Solution

Question 10

\(\sqrt{2} \mathrm{R}_{\mathrm{e}}\)Re

\(\frac{R_{e}}{\sqrt{2}}\)

Re

\(\frac{\sqrt{2}}{R_{e}}\)

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.

The time period of S₁ is four times that of S₂.

S₁ and S₂ are moving with the same speed.

\(\sqrt{2} \mathrm{R}_{\mathrm{e}}\)R_e

R_e