cbse 11th physics | Online Objective Test

Q1. A steam engine delivers 5.4

\times

1 0^{8}

J of work per minute and services

3.6 \times 10^{9} J

of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

A. 12 percent, 2.8

\times

1 0^{9}

B. 14 percent, 3.0

\times

1 0^{9}

C. 15 percent, 3.0

\times

1 0^{9}

D. 13 percent, 2.9

\times

1 0^{9}

Answer : Option C
Explaination / Solution:

$\begin{aligned} η = \frac{W o r k D o n e}{H e a t E x t r a c t e d} = \frac{5.4 \times 10^{8}}{3.6 \times 10^{9}} = 0.15 = 15 % \\ H e a t w a s t e d = H e a t E x t r a c t e d - W o r k D o n e \end{aligned}$

Heat Wasted = $3.6 \times 10^{9} - 5.4 \times 10^{8} = 3.06 \times 10^{9} J$

Laws of Motion Prepare / Learn

Q2.

A truck starts from rest and accelerates uniformly at 2.0 m $s^{- 2}$ . At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in m $s^{- 2}$ ) of the stone at t = 11s? (Neglect air resistance.)

A. 14

B. 12

C. 8

D. 10

Answer : Option D
Explaination / Solution:

When the stone is dropped from the truck , the horizontal force acting on it become zero after stone is released. Stone continue to move under the influence of gravity only. So that acceleration of the stone is equal to the gravitational acceleration g.

a = g = 10ms-2 acting vertically downward.

Work Energy and Power Prepare / Learn

Q3. A 6.0-kg block initially at rest is pulled to the right along a horizontal, surface having a coefficient of kinetic friction of 0.15, by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.

A. 1.8 m/s

B. 0.8 m/s

C. 2.8 m/s

D. 3.8 m/s

Answer : Option A
Explaination / Solution:

ΔK=W

F=12N

f=μR=μmg=0.15×6×9.8=8.82N

12mv2−12mu2=Fnets

(12×6×v2)−0=3.18×3

3v2=3.18×3

v2=3.18

v=1.8m/sec

Physical World Prepare / Learn

Q4. Albert Einstein was awarded the Nobel Prize for his work on

A. Explanation of photoelectric effect

B. E =

m c^{2}

C. scattering of light

D. Theory of relativity

Answer : Option A
Explaination / Solution:

The Nobel Prize in Physics 1921 was awarded to Albert Einstein "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect". Albert Einstein received his Nobel Prize one year later, in 1922.

System of Particles and Rotational Motion Prepare / Learn

Q5. A jet airplane travelling at the speed of 500 kmh–1 ejects its products of combustion at the speed of 1500 km h–1relative to the jet plane. What is the speed of the combustion with respect to an observer on ground?

A. -2550 km/h

B. -2000 km/h

C. -1250 km/h

D. -1000 km/h

Answer : Option D
Explaination / Solution:

Speed of the jet airplane, vjet = 500 km/h Relative speed of its products of combustion with respect to the plane, vsmoke = – 1500 km/h

Speed of its products of combustion with respect to the ground = v′smoke

Relative speed of its products of combustion with respect to the airplane, vsmoke = v′smoke – vjet

=> – 1500 = v′smoke – 500

=> v′smoke = – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

Mechanical Properties of Fluids Prepare / Learn

Q6. If two soap bubbles of different radii are in communication with each other

A. air flows from the smaller bubble into the larger one and large bubble grows at the expense of the smaller one

B. air flows from the larger into the smaller bubble until the radius of the smaller one becomes half that of the larger one

C. air flows from the larger bubble into the smaller one until the two bubbles are of equal size

D. the sizes of the bubbles remain unchanged

Answer : Option A
Explaination / Solution:
No Explaination.

Kinetic Theory Prepare / Learn

Q7. Nine particles have speeds of 5.00, 8.00, 12.0, 12.0, 12.0, 14.0, 14.0, 17.0, and 20.0 m/s. find the particles’ average speed.

A. 16.7 m/s

B. 8.7 m/s

C. 12.7 m/s

D. 10.7 m/s

Answer : Option C
Explaination / Solution:

vav=∑vn=5.0+8.0+12.0+12.0+12.0+14.0+14.0+17.0+20.09vav=12.7m/sec

Thermal Properties of Matter Prepare / Learn

Q8. The triple points of carbon dioxide is 216.55 K. Express this temperatures on the Celsius and Fahrenheit scales

A. – 66.60

^{\circ}

C = – 99.88

^{\circ}

B. – 56.60

^{\circ}

C = – 69.88

^{\circ}

C. – 76.60

^{\circ}

C = – 120.88

^{\circ}

D. – 86.60

^{\circ}

C = – 150.88

^{\circ}

Answer : Option B
Explaination / Solution:

$\begin{aligned} \frac{T_{C} - 0}{100 - 0} = \frac{T_{F} - 32}{212 - 32} = \frac{T_{K} - 273.15}{373.15 - 273.15} \\ \frac{T_{C}}{100} = \frac{T_{F} - 32}{180} = \frac{T_{K} - 273.15}{100} \end{aligned}$

TK = 216.55K

after solving

TC = – 56.60 $^{\circ}$ C

TF = – 69.88 $^{\circ}$ F

Units and Measurements Prepare / Learn

Q9. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

A. 0.025 mm

B. 0.035 mm

C. 0.040 mm

D. 0.030 mm

Answer : Option B
Explaination / Solution:

Magnification of microscope = 100 Observed width of the hair = 3.5 mm Estimate on the thickness of hair is given by

Magnification = $\frac{O b s e r v e d w i d t h}{R e a l w i d t h}$

=> Real width = $\frac{O b s e r v e d w i d t h}{M a g n i f i c a t i o n}$ = $\frac{3.5}{100}$ = 0.035 mm

Oscillations Prepare / Learn

Q10. For all practical purposes, the motion of a simple pendulum is SHM,

A. Only if the maximum angle which the string makes with the vertical is less than I

B. Only if the length of its string is at least one meter

C. Only if the maximum angle which its string makes with the vertical is less than 342.

D. None of these

Answer : Option D
Explaination / Solution:
No Explaination.

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