Physics Test

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Physics Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Represent 1 kWh in terms of joules (J).

A

B

C

D
None of these

Solution
Question 2
1 / -0

The smallest unit that is used to express the size of an atom is

A
parsec

B
light year

C
angstrom

D
nanometre

Solution

The smallest unit that is used to express the size of an atom is angstrom.
Question 3
1 / -0

Which of the following options is not a unit of length?

A
Angstrom

B
Light year

C
Fermi

D
Radian

Solution

A radian is a unit of measurement for angles defined by the ratio of the length of the arc of a circle to the radius of that circle. One radian is the angle at which that ratio equals one
Question 4
1 / -0

What is the SI unit of pressure?

A
Nm^–2

B
Nm

C
Nm^–1

D
Nm²

Solution

For pressure, the SI system's basic unit is Pascal (Pa), which is N/m² (New tonper squaremeter, while Newtonis kgf/s²).
Question 5
1 / -0

The density of the material of a cube is determined by measuring its mass and edge length. If the maximum errors in the measurement of the mass and edge length are 3% and 2%, respectively, then the maximum error in the measurement of density is

A
1%

B
5%

C
7%

D
9%

Solution

Density \(d=\frac{\text {Mass}}{\text {Volume}}\)
Or \(d=\frac{m}{a^{3}}\)
Percentage error in density \(\% d=\% m+3 \times \% a\)
\(\Longrightarrow \% d=3+3 \times 2=9 \%\)
Question 6
1 / -0

Ampere-hour is the unit of

A
electrical energy

B
electric potential

C
electric current

D
electric charge

Solution

Ampere-hour is the unit ofelectric charge
Question 7
1 / -0

A force F is applied horizontally to a block A of mass m₁ which is in contact with a block B of mass m₂ as shown in the figure. If the surfaces are frictionless, the force exerted by A on B is given by:

A
\(\frac{m_{1}}{m_{2}} F\)

B
\(\frac{m_{2}}{m_{1}} F\)

C
\(\frac{m_{1} F}{\left(m_{1}+m_{2}\right)}\)

D
\(\frac{m_{2} F}{\left(m_{1}+m_{2}\right)}\)

Solution

Let the acceleration of the whole system be \(a\) and the normal force acting between the blocks be \(N\)
Total mass of the system \(m=m_{1}+m_{2}\)
\(\therefore\) Acceleration of the system \(a=\frac{F}{m}=\frac{F}{m_{1}+m_{2}}\)
Thus normal force between the blocks \(N=M_{2} a=\frac{m_{2} F}{m_{1}+m_{2}}\)
Question 8
1 / -0

A force acts for 10 s on a body of mass 100 kg, after which the force ceases to act and the body describes 160 m in the next 8 s. What is the magnitude of the force? (Assume no frictional losses)

A
200 N

B
250 N

C
300 N

D
400 N

Solution

Say initial velocity is 0 then,
\(v=u+a t\)
\(v=0+a 10\)
\(\mathrm{v}=10 \mathrm{a}\)
After 10 s body starts moving and cover \(100 \mathrm{m}\) in \(5 \mathrm{s}\)
\(S=v t\)
\(100=(10 a) 5\)
\(a=2 m / s 2\)
using the value of a in \(\mathrm{F}=\mathrm{ma}\)
\(\mathrm{F}=100 \times 2=200 \mathrm{N}\)
Question 9
1 / -0

The pulleys and the strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, angle θ should be

A
0°

B
30°

C
45°

D
60°

Solution

For block
\(A, T-m g=m(0)\)
For block \(\mathrm{B}, \mathrm{T}-\mathrm{mg}=\mathrm{m}(\mathrm{O})\)
(since the both \(A\) and \(B\) blocks are similar and tension in the strings are \(m g\). It has zero acceleration, as system is in equilibrium).
So, \(2 T \cos \theta=\sqrt{2} m g\)
\(2 m g \cos \theta=\sqrt{2} m g\)
\(\theta=45^{\circ}\)
Question 10
1 / -0

Directions: The following question has four choices, out of which ONLY ONE is correct.
A particle moves in the xy-plane under the action of a force F, such that the value of its linear momentum p at any time t is p_x= 2 cos t, p_y = 2 sin t. The angle between F and p at any given time t will be

A
90^o

B
0^o

C
180^o

D
30^o

Solution

From Newton's second law,
Force \(F\) is nothing but the rate of change of momentum.
Hence, \(F=d P / d t\)
Given, \(P_{x}=2 \operatorname{cost} \& P_{y}=2 \sin t\)
\(\Rightarrow \mathrm{F}_{\mathrm{x}}=-2 \operatorname{sint} \& \mathrm{F}_{\mathrm{y}}=2 \mathrm{cost}\)
Now to find the angle between the force vector and the momentum vector, let us first find the dot product of these two vectors.
\(\mathrm{F} . \mathrm{P}=\mathrm{F}_{\mathrm{x}} \mathrm{P}_{\mathrm{x}}+\mathrm{F}_{\mathrm{y}} \mathrm{P}_{\mathrm{y}}\)
\(=2 \cos t\times-2 \sin t+2 \sin t \times 2 \cos t\)
\(=-4 \cos t \times \sin t+4 \cos t \times \sin t\)
\(=0\)
Also, \(\mathrm{F} . \mathrm{P}=\mathrm{|F} \| \mathrm{P} \mid \cos \theta\)
\(\Rightarrow \cos \theta=0\)
\(\Rightarrow \theta=90\) degrees.
Hence, the angle between force and momentum vectors is 90 degrees