cbse 11th physics | Online Objective Test

Motion in A Plane Prepare / Learn

Q1. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30

^{\circ}

, what is the speed of the aircraft?

A. 172 m/s

B. 192 m/s

C. 193 m/s

D. 182 m/s

Answer : Option D
Explaination / Solution:

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ = 30°
Time = 10 s
In ΔPRO:
tan 15° = $\frac{P R}{O R}$
PR = OR tan 15°
= $3400 \times t a n 15 °$

ΔPRO is similar to ΔRQO.
∴PR = RQ
PQ = PR + RQ
= 2PR = 2 × 3400 tan 15°
= 6800 × 0.268 = 1822.4 m
∴ Speed of the aircraft = $\frac{1822.4}{10}$

= 182.24 m/s $\approx 182$ m/s

Mechanical Properties of Solids Prepare / Learn

Q2.

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5

\times

1 0^{- 3}

c m^{2}

in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the tensile strain.

A. 3.3

\times

1 0^{- 3}

upper, 2.4

\times

1 0^{- 3}

lower

B. 3.4

\times

1 0^{- 3}

upper, 2.5

\times

1 0^{- 3}

lower

C. 3.2

\times

1 0^{- 3}

upper, 2.3

\times

1 0^{- 3}

lower

D. 3.1

\times

1 0^{- 3}

upper, 2.0

\times

1 0^{- 3}

lower

Answer : Option D
Explaination / Solution:
No Explaination.

Thermodynamics Prepare / Learn

Q3.

A 1.0-mol sample of an ideal gas is kept at 0.0 $^{\circ}$ C during an expansion from 3.0 L to 10.0 L. How much energy transfer by heat occurs with the surroundings in this process?

A. 2.8

\times

1 0^{3}

B. 2.9

\times

1 0^{3}

C. 2.7

\times

1 0^{3}

D. 3.0

\times

1 0^{3}

Answer : Option C
Explaination / Solution:

in isothermal process $Δ U = 0$

hence $Δ Q = W$

$Q = n R T \log_{e} \frac{V_{f}}{V_{i}}$

$Q = 1 \times 8.31 \times 273 \times 2.303 \log_{10} \frac{10}{3} = 2.7 \times 10^{3} J$

Laws of Motion Prepare / Learn

Q4. A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m

s^{- 1}

, what would be the reading on the scale?

A. 35 kg

B. 75 kg

C. 105 kg

D. 70 kg

Answer : Option D
Explaination / Solution:

Mass of the man, m = 70 kg

Acceleration, a = 0

UsingNewton’s second law of motion,

wecan write the equation of motion as:

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

∴ R = mg = 70 × 10 = 700 N

∴ Reading on the weighing scale = 700 / g = 700 / 10 = 70 kg

Work Energy and Power Prepare / Learn

Q5. A body of mass 0.5 kg travels in a straight line with velocity v =a

x^{3 / 2}

where a = 5

m^{- 1 / 2} s^{- 1}

. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

A. 60 J

B. 50 J

C. 40 J

D. 30 J

Answer : Option B
Explaination / Solution:

Mass of the body, m= 0.5 kg

Velocity of the body $v = a x^{3 / 2}$

$a = 5 m^{- 1 / 2} s^{- 1}$

Initial velocity at x = 0 is u = 0

Final velocity at x = 2 m is $v = 10 \sqrt{2} m / s$

work done = Change in kinetic energy

$\begin{aligned} W = K_{f} - K_{i} \\ W = \frac{1}{2} m v^{2} - 0 \\ W = \frac{1}{2} \times 0.5 \times {(10 \sqrt{2})}^{2} = \frac{1}{2} \times 0.5 \times 200 = 50 J \end{aligned}$

Physical World Prepare / Learn

Q6. Which of these is not a fundamental force?

A. Electromagnetic Force

B. Frictional force

C. Strong Nuclear Force

D. Gravitational Force

Answer : Option B
Explaination / Solution:

There are four conventionally accepted fundamental interactions—gravitational, electromagnetic, strong, and weak. Each one is described mathematically as a field.

System of Particles and Rotational Motion Prepare / Learn

Q7. A body of mass 2.0 kg and radius of gyration 0.5 m is rotating about an axis. If its angular speed is 2.0 rad/s, the angular momentum of the body (in kg -

m^{2}

/s) is

A. 2.0

B. 0.5

C. 1.5

D. 1.0

Answer : Option D
Explaination / Solution:

$\begin{aligned} I = m k^{2} \\ m = 2 K g \\ k = 0.5 m \\ I = 2 \times 0.5 \times 0.5 = 0.5 K g m^{2} \end{aligned}$

angular momentum

$\begin{aligned} L = I ω \\ ω = 2 r a d / s \\ L = 0.5 \times 2 = 1.0 K g m^{2} / s \end{aligned}$

Mechanical Properties of Fluids Prepare / Learn

Q8. A drop of oil is placed on the surface of water. Which of the following statements is correct?

A. It will spread as a thin layer

B. It will partly changed to spherical droplets and partly as thin film

C. It will remain on it as a sphere

D. It will float as a distorted drop on the water surface

Answer : Option A
Explaination / Solution:

when any liquid is poured onto another liquid there are different phenomena that can take place depending on the properties of liquids:In case of crude oil and water - The two liquids are non-miscible and the first is lighter than the second. The first liquid then spreads on the surface of the second liquid forming a thin layer. The thickness of this layer depends on the properties of the two liquids. This is the case for most hydrocarbons, including both refined products and crude oil of API gravity greater than 10.

Kinetic Theory Prepare / Learn

Q9. In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600

m^{2}

at an angle of 45.0

^{\circ}

to the window surface. Each hailstone has a mass of 5.00 g and a speed of 8.00 m/s. If the collisions are elastic, what is the pressure on the window?

A. 1.57 Pa

B. 1.97 Pa

C. 1.32 Pa

D. 1.13 Pa

Answer : Option A
Explaination / Solution:

P=FAF=500×ΔpΔt=500×2mvcos45∘Δt=500×2×5×10−3×8×130×1.41A=0.6m2P=500×2×5×10−3×8×130×1.41×0.6=1.57Pa

Thermal Properties of Matter Prepare / Learn

Q10.

A steel tape 1m long is correctly calibrated for a temperature of 27.0 $^{\circ}$ C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 $^{\circ}$ C. What is the actual length of the steel rod on that day? Coefficient of linear expansion of steel = 1.20 $\times$ $1 0^{- 5}$ $K^{- 1}$

K^{- 1}

A. 63.0126 cm

B. 63.0136 cm

C. 63.0146 cm

D. 63.0116 cm

Answer : Option B
Explaination / Solution:

$Δ l = α l Δ T = 1.20 \times 10^{- 5} \times 63 \times 18 = 0.0136 c m$

final length =63.0136cm

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