M.C.Q (1 Marks)

MCQ 11 Mark

A tank with a square base of area $2.0 m^2$ is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area $20 cm^2$ at the bottom of the partition. Water is filled in one compartment and acid, of relative density 1.7 , in the other, both to a height of 4 m . If $g =10 ms^{-2}$ the force necessary to keep the door closed is:

A
80.88 N
B
20.92 N
C
40.52 N
D
54.88 N

Answer

(d) 54.88 N
Explanation: Water compartment
$
\begin{aligned}
& P=h \rho g \\
& =4 \times 1.0 \times 10^3 \times 9.8 \\
& =39.2 \times 10^3 Pa
\end{aligned}
$
Acid Compartment,
$
\begin{aligned}
& P^{\prime}=h \rho^{\prime} g \\
& =4 \times 1.7 \times 10^3 \times 9.8 \\
& =66.64 \times 10^3 Pa
\end{aligned}
$
Now,
$
\begin{aligned}
& P^{\prime}-P=66.64 \times 10^3-39.2 \times 10^3 \\
& P^{\prime}-P=27.44 \times 10^3 Pa \\
& A=20 cm^2=20 \times 10^{-4} m^2
\end{aligned}
$
Force using pressure,
$
\text { Pressure }=\frac{\text { Force }}{\text { Area }}
$
Force $=$ pressure $\times$ area $F =27.44 \times 10^3 \times 20 \times 10^{-4}$
$
F=54.88 N
$

View full question & answer→

MCQ 21 Mark

Measurement of a physical quantity is essentially the

A
process of comparing with a standard using an instrument
B
process of observing the physical quantity
C
process of taking readings on an instrument
D
process of subdividing the physical quantity

Answer

(a) process of comparing with a standard using an instrument
Explanation: The Measurement of a given quantity is essentially an act or result of comparison between a quantity whose magnitude (amount) is unknown, with a similar quantity whose magnitude (amount) is known, the latter quantity being called a Standard.

View full question & answer→

MCQ 31 Mark

A liquid boils when its vapour pressure is equal to

A
atmospheric pressure
B
double of atmospheric pressure
C
6.0 cm of Hg column
D
1000 pa or more

Answer

(a) atmospheric pressure
Explanation: When vapour pressure is equal to atmospheric pressure, then boiling occurs.

View full question & answer→

MCQ 41 Mark

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2 t^3-6 t^2$. The torque on the wheel becomes zero at

A
$t =0.25 s$
B
$t =0.5 s$
C
$t=2 s$
D
$t=1 s$

Answer

(d) $t=1 s$
$
\begin{aligned}
& \text { Explanation: } \theta(t)=2 t^3-6 t^2 \\
& \omega=\frac{d \theta}{d t}=6 t^2-12 t \\
& \alpha=\frac{d \omega}{d t}=12 t-12=0 \\
& \Rightarrow t=1 s
\end{aligned}
$

View full question & answer→

MCQ 51 Mark

A closed organ pipe (closed at one end) is excited so as to support the third overtone. It is then found that there are in the pipe:

A
three nodes and four antinodes
B
four nodes and three antinodes
C
four nodes and four antinodes
D
three nodes and three antinodes

Answer

(c) four nodes and four antinodes
Explanation:In third overtone, the length of the closed organ pipe is $\frac{7 \lambda}{4}$. Air in the pipe will have four nodes and four antinodes as shown in the figure.

View full question & answer→

MCQ 61 Mark

The wavelength difference of light waves of wave numbers $2 \times 10^6 / m$ and $2.25 \times 10^6 / m$ is

A
$0.556 \times 10^6 m$
B
$0.556 \times 10^{-6} m$
C
$0.0556 \times 10^6 m$
D
$0.0556 \times 10^{-6} m$

Answer

$\begin{aligned} & \text { } (d) 0.0556 \times 10^{-6} m \\ & \text { Explanation: } \Delta \lambda=\lambda_1-\lambda_2=\frac{1}{2 \times 10^6}-\frac{1}{2.25 \times 10^6} \\ & =\frac{0.25 \times 10^{-6}}{2 \times 2.25} m=0.0556 \times 10^{-6} m\end{aligned}$

View full question & answer→

MCQ 71 Mark

A jet lands on an aircraft carrier at 30 m/s. It stops in 2.0 s. What is the displacement of the plane when it stops?

A
45
B
30
C
35
D
40

Answer

(b) 30
Explanation: Initial velocity u, $=30 m / s$
As it stop so final velocity, $v =0 m / s$
Time $t =2 s$
Distance covered $= s$
we know that, $v = u + at$
$
\begin{aligned}
& 0=30+a(2) \\
& a=-15 m / s^2
\end{aligned}
$
We know, $s =u t+\frac{1}{2} a t^2$
$
\begin{aligned}
& \Rightarrow s=30 \times 2+\frac{1}{2}(-15) \times 2^2 \\
& \Rightarrow s=30 m
\end{aligned}
$

View full question & answer→

MCQ 81 Mark

Deimos, a moon of Mars, is about 12 km in diameter with mass $2.0 \times 10^{15} kg$. Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it?

A
4.7 m/s
B
5.1 m/s
C
4.3 m/s
D
4.9 m/s

Answer

(a) $4.7 m / s$
Explanation: $v=\sqrt{\frac{G M}{r}}$
Here, $G =6.67 \times 10^{-11} N m ^2 kg^{-2}$
$
\begin{aligned}
& M=2 \times 10^{15} kg \\
& r=\frac{12000}{2}=6000 \\
& \Rightarrow v=\sqrt{\frac{6.67 \times 10^{-11} \times 2 \times 10^{15}}{6000}} \\
& \Rightarrow v=4.7 m / sec
\end{aligned}
$

View full question & answer→

MCQ 91 Mark

A particle falls towards earth from infinity. Its velocity on reaching the earth would be:

A
Infinity
B
Zero
C
$2 \sqrt{g R}$
D
$\sqrt{2 g R}$

Answer

(d) $\sqrt{2 g R}$
$
\begin{aligned}
& \text { Explanation: } \frac{1}{2} m v^2=U_i-U_f=0-\left(-\frac{G M m}{R}\right) \\
& \frac{1}{2} m v^2=\frac{g R^2 m}{R}\left[GM=gR^2\right] \\
& v=\sqrt{2 g R}
\end{aligned}
$

View full question & answer→

MCQ 101 Mark

What is the distance of the centre of mass of a half ring from its centre if its radius is 0.5 m?

A
$\frac{1}{3 \pi} m$
B
$\frac{1}{\pi} m$
C
$\frac{2}{3 \pi} m$
D
$\frac{1}{2 \pi} m$

Answer

(b) $\frac{1}{\pi} m$
Explanation: Distance of the centre of mass of a half ring from its centre, $y =\frac{2 R}{\pi}=\frac{2 \times 0.5}{\pi} m=\frac{1}{\pi} m$

View full question & answer→

MCQ 111 Mark

Equation of plane wave is $y=4 \sin \frac{\pi}{4}\left[2 t+\frac{x}{8}\right]$
The phase difference at any given instant of two particles 16 cm apart is

A
$60^{\circ}$
B
$30^{\circ}$
C
$90^{\circ}$
D
$120^{\circ}$

Answer

(c) $90^{\circ}$
$
\begin{aligned}
& \text { Explanation: } y=4 \sin \left[\frac{2 \pi}{4} t+\frac{\pi}{32} x\right] \\
& y=A \sin \left[\frac{2 \pi}{T} t+\frac{2 \pi}{\lambda} x\right] \\
& \Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{\pi}{32} \times 16=\frac{\pi}{2}
\end{aligned}
$

View full question & answer→

MCQ 121 Mark

A raindrop is released from a cloud 1000 m above ground. When the drop is about to hit the ground, its speed will be

A
Cannot be predicted
B
Constant terminal speed
C
Decreasing due to retardation from air drag
D
Increasing due to acceleration due to gravity

Answer

(b) Constant terminal speed
Explanation: The rain drops acquire terminal velocity after falling long distance due to balance of buoyant force and weight of the drop due to gravity.

View full question & answer→

Physics M.C.Q (1 Marks)

Take a timed test