cbse 11th physics | Online Objective Test

Q1.

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\overset{0}{A} : 1 \overset{0}{A} = 1 0^{- 10}$ m. The size of a hydrogen atom is about $0. 5 \overset{0}{A}$ . What is the total atomic volume in $m^{3}$ of a mole of hydrogen atoms ?

A. ≈3×10−7m3

B. ≅3.2×10−6m3

C. ≅3.3×10−7m3

≅ 2.9 \times 10^{- 6} m^{3}

≅ 2.9 \times 10^{- 6} m^{3}

Answer : Option A
Explaination / Solution:

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10-10 m
Volume of hydrogen atom = (4/3) π r3
= (4/3) × (22/7) × (0.5 × 10-10)3
= 0.524 × 10-30 m3
1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3

$\approx 3 \times 10^{- 7} m^{3}$

Oscillations Prepare / Learn

Q2. A cylinder floats vertically in a tank full of water. The period of oscillation of the cylinder when depressed and released is dependent on

A. Density of the liquid

B. The mass of the cylinder

C. Length of cylinder immersed in the liquid at equilibrium

D. The area of cross section of the cylinder

Answer : Option D
Explaination / Solution:
No Explaination.

Motion in A Straight Line Prepare / Learn

Q3.

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the maximum height it travels in meters. g =9.8 m / $s e c^{2}$

A. 30.4

B. 15.4

C. 20.4

D. 25.4

Answer : Option C
Explaination / Solution:

Initial velocity u = 20.0 m/s

At maximum height stone ll be stopped,

So final velocity v = 0 m/s

Acceleration due to gravity a = g =

-9.8 m/s2 (-ve Because it is in opposite direction of motion)

Let maximum height = s

We know,

$v^{2} - u^{2} = 2 a s$

=> $0^{2} - (20)^{2} = 2 \times (- 9.8) s$

$=> - 400 = - 19.6 s$

$=> s = \frac{- 400}{- 19.6} = 20.4 m$

Waves Prepare / Learn

Q4. Longitudinal waves cannot be propagated through

A. vacuum

B. a solid

C. a gas

D. a liquid

Answer : Option A
Explaination / Solution:
No Explaination.

Motion in A Plane Prepare / Learn

Q5.

At a distance L=400m from the traffic lights the brakes are applied to a locomotive moving at a velocity v=54km/hr. Determine the position of locomotive relative to the traffic lights 1 minute after the application of the brakes if its acceleration is -0.3m/s2.

A. 15 m

B. 25 m

C. 30 m

D. 40 m

Answer : Option B
Explaination / Solution:

Initial velocity u = 54 km/h = 15 m/s

Let Final velocity v = 0

Acceleration a = -0.3 m/s2

Time taken to stop = t

Using $v = u + a t$

=> $0 = 15 + (- 0.3) t$

=> t = 50 sec

It means it has been stopped before 1 minutes.

So distance covered in 1 minute

$v^{2} - u^{2} = 2 a s$

=> $0 - 225 = 2 \times (- 0.3) s$

=> s = 375 m

Position of locomotive relative to the traffic lights= 400-375 = 25 m

Mechanical Properties of Solids Prepare / Learn

Q6.

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013

\times

1 0^{5}

Pa), Final volume = 100.5 litre

A. 2.126

\times

1 0^{9}

B. 2.226

\times

1 0^{9}

C. 2.026

\times

1 0^{9}

D. 2.326

\times

1 0^{9}

Answer : Option C
Explaination / Solution:
No Explaination.

Thermodynamics Prepare / Learn

Q7.

Consider the cyclic process depicted in Figure If Q is negative for the process BC, and if $Δ E_{i n t}$ is negative for the process CA, what are the signs of W and $Δ E_{i n t}$ that are associated with BC?

A. 0, +

B. 0, -

C. 0, -

D. +, -

Answer : Option C
Explaination / Solution:

W = area under P-V curve = 0

$\begin{aligned} Δ Q = Δ E_{int} + W \\ W = 0, Δ Q = - v e \\ Δ E_{int} = - v e \end{aligned}$

Laws of Motion Prepare / Learn

Q8.

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m $s^{- 1}$ , what is the recoil speed of the gun?

A. 1.8 cm/s

B. 1.6 cm

s^{- 1}

C. 1.7 cm/s

D. 1.9 cm/s

Answer : Option B
Explaination / Solution:

Massof shell, $m = 0.02 k g$

Mass of gun, $M = 100 k g$

Speed of shell, $v = 80 m s^{- 1}$

Let V be the recoil speed of the gun. According to the law of conservation of momentum,

Initial momentum = Final momentum

$0 = m v + M V V = - \frac{m v}{M} = - \frac{0.02 \times 80}{100} V = - 0.016 m s^{- 1} = - 1.6 c m s^{- 1}$

Negative sign indicates that the gun moves backward as the bullet moves forward.

Work Energy and Power Prepare / Learn

Q9. A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity

A. 30 J, 30 J

B. 25 J, -25 J

C. 0, 0

D. 10 J, 10 J

Answer : Option C
Explaination / Solution:

Work done by conservative force in round trip will be zero (displacement = 0). So that work done by tension and gravity will be zero.

System of Particles and Rotational Motion Prepare / Learn

Q10. A particle is orbiting in a vertical plane. Its linear momentum will be directed

A. horizontally

B. at 45

^{\circ}

to the vertical

C. vertically

D. tangential to the orbit.

Answer : Option D
Explaination / Solution:

As the direction of velocity at any point moving in circular orbit is tangent at that point, thus momentum would be tangential to the orbit.

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