CBSE Sample Papers For Class 11 Physics Mock Paper 1

Q1. The dimension of energy is ________ which is also the dimension of ____________.

Opt:

MLT, force

MLT^-1, pressure

MLT^-2, power

ML²T^-2, work

Ans: ML²T^-2, work

Q2. A ball is dropped downwards, after 1 second another ball is dropped downwards from the same point. The distance between them after 3 seconds will be

Opt:

20 m

25 m

50 m

98 m

Ans: 25 m

Q3. A body A moves with a uniform acceleration ‘a’ and zero initial velocity. Another body B starting from the same point moves in the same direction with a constant velocity v. The two bodies will meet after a time

Opt:

3a/2v

v/2a

v/3a

2v/a

Ans: 2v/a

Q4. The range of a particle when launched at an angle of 15^o with the horizontal is 1.5 km. The range of the projectile when launched at an angle of 45^o to the horizontal is

Opt:

1.5 km

2 km

2.5 km

3 km

Ans: 3 km

Q5. A wooden block of mass M is placed on a smooth inclined plane of inclination 60° with the horizontal. The force exerted by the plane on the wooden block has a magnitude

Opt:

0.5Mg

1.0Mg

1.5Mg

2.0Mg

Ans: 0.5Mg

Q6. A ball is dropped from a height of 1m. If coefficient of restitution between the surface and the ball is 0.6, the ball rebounds to a height of

Opt:

1 m

0.6 m

0.4 m

0.36 m

Ans: 0.36 m

Q7. A meter stick is balanced of a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark the stick is found to be balanced at 45.0 cm. What is the mass of metre stick?

Opt:

55 g  45 g  50 g  66 g

Ans: 66 g

Q8. Two satellites of mass 4m and m respectively, orbit the earth in circular orbits of radii r and 4r. The ratio of their orbital speeds is

Opt:

4:1

3:1

2:1

1:1

Ans: 2:1

Q9. If one mole of a monatomic gas (γ= 5/3) is mixed with one mole of a diatomic gas (γ = 7/5), the value of γ for the mixture is

Opt:

3.07

1.53

1.50

1.40

Ans: 1.50

Q10. A manometer connected to a closed tap reads 450000 Pa. When the tap is opened, the reading of manometer falls to 400000 Pa. Then the velocity of flow of water is

Opt:

7 ms^-1

8 ms^-1

9 ms^-1

10 ms^-1

Ans: 10 ms^-1

Q11. An elastic material with Young’s modulus ‘Y’ is subjected to a tensile stress ‘S’, elastic energy stored per unit volume of the material is

Opt:

YS/2

S²/Y

S²/2Y

S/2Y

Ans: S²/2Y

Q12. A point source emits waves in all directions in a non-absorbing medium. The ratio of the amplitudes of waves at two points distant 9 and 25 m from the source is

Opt:

9:25

25:9

3:5

5:3

Ans: 25:9

Q13. The most appropriate material for cooking pot is the one having

Opt:

high specific heat and low conductivity

high specific heat and high conductivity

low specific heat and high conductivity

low specific heat and high conductivity

Ans: low specific heat and high conductivity

Q14. For an ideal gas of pressure P and volume V,

Opt:

adiabatic curve is always less steeper than isothermal curve.

adiabatic curve is always steeper than isothermal curve.

adiabatic curve has same steep as isothermal curve.

adiabatic curve always coincides with isothermal curve.

Ans: adiabatic curve is always steeper than isothermal curve.

Q15. A particle has displacement y given by y = 3 sin (5πt + ∅) where y is in metre and t is in seconds. The frequency and period of motion is

Opt:

0.4 Hz, 2.5 s

2.5 Hz, 0.4 s

2.5 Hz, 2.5 s

0.4 Hz, 0.4 s

Ans:

2.5 Hz, 0.4 s

Q16. The ratio of speed of A and B from the graph is

Opt:

1 : 1

1 : 2

1 : 3

1 : 4

Ans: 1 : 3

Q17. A steel wire of 10 m in length is stretched through 5 mm. If Young’s modulus of steel is 2×10¹¹ N/m², what is the energy density of wire?

Opt:

2.5×10⁴ J/m²  1.5×10³ J/m²  4.5×10³ J/m²  4×10³ J/m²

Ans: 2.5×10⁴ J/m²

Q18. Let k be the force constant of a spring. The work done in increasing its length from l₁ to l₂ will be

Opt:

k (l₁ – l₂)

k(l₁ + l₂)/2

k (l₁ ²  + l₂²)

k (l₁ ²  – l₂²)/2

Ans: k (l₁ ²  – l₂²)/2

Q19. Calculate the escape velocity from the moon. It is given that mass of the moon = 7.4 X 10²² kg and radius of the moon is 1740 km.

Ans: The escape velocity is given by

Q20. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Ans:

Let, mass of the meter stick= m’Here, mass of each coin, m=5 gAs the coins are placed 12 cm away from the end P, the centre of mass gets displaced by 5 cm from point R towards the end P.The centre of mass is positioned at the seperation of 45 cm from point P.For the equilibrium state about the 45 cm mark, we have:10×g(45−12)=m‘g(50−45)∴m‘=10×335=66g∴Mass of the metre stick, m’ =66 g.

Q21. It is required to prepare a steel meter scale, such that the millimeter intervals are to be accurate within 0.0005mm at a certain temperature. Determine the maximum temperature variation allowable during the rulings of millimeter marks.

OR

A refrigerator is to maintain eatables kept at 9⁰C. If room temperature is 36⁰C, calculate the coefficient of performance.

Ans:

Given

for steel = 1.322 x  10^{-5 0}C^-1​

 

OR

Here T₁ = 36⁰C = 36 + 273 = 309 k

 T₂ = 9⁰C = 9 + 273 = 282 k

Coefficient of performance

 = T₂ / (T₁ – T₂)

= 282 / (309 – 282)

= 282/ 27

= 10.4

Q22. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^-1 can go without hitting the ceiling of the hall?

Ans:

\begin{array}{l}\text{Given,}\text{speed}\text{of}\text{ball,}\text{u = 40}{\text{ms}}^{\text{-1}}\\ \text{Maximum}\text{height,}\text{h = 25}\text{m}\\ \text{Let}\text{θ}\text{be}\text{the}\text{angle}\text{of}\text{projection}\text{with}\text{the}\text{horizontal}\text{.}\\ \text{For}\text{projectile}\text{motion,}\\ \text{Maximum}\text{height,}\text{H = }\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\text{θ}}{\text{2g}}\\ \text{25 m = }\frac{{\left({\text{40 ms}}^{\text{-1}}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\text{θ}}{\text{2×9}{\text{.8 ms}}^{\text{-2}}}\\ \therefore {\text{sin}}^{\text{2}}\text{θ = 0}\text{.30625}\\ \text{sinθ = 0}\text{.5534}\\ \therefore {\text{θ = sin}}^{\text{-1}}\left(\text{0}\text{.5534}\right)\text{ = 33}\text{.60}°\\ \text{Horizontal}\text{range,}\text{R = }\frac{{\text{u}}^{\text{2}}\text{sin2θ}}{\text{g}}\\ \text{R = }\frac{{\left({\text{40 ms}}^{\text{-1}}\right)}^{\text{2}}\text{×sin2×33}\text{.6}°}{\text{9}{\text{.8 ms}}^{\text{-2}}}\text{ = 150}\text{.5}\text{m}\end{array}

Q23. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^–1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^–1.

Ans:

\begin{array}{l}\text{Here,}\text{operating frequency of the SONAR},\nu =\text{4}0\text{ kHz}\\ \text{Speed of the}\text{enemy submarine,}{\text{v}}_{\text{e}}=\text{36}0{\text{ kmh}}^{\text{-1}}\\ =\text{ 1}00{\text{ ms}}^{\text{-1}}\\ \text{Speed of sound in the}\text{water,}\text{v}\text{=}{\text{1450 ms}}^{\text{-1}}\\ \text{As}\text{the source is at rest and the observer}\text{(enemy}\text{submarine) is moving toward it,}\\ \therefore \text{Apparent frequency }\left(\nu \text{‘}\right)\text{ received and reflected by}\text{the submarine is given}\text{as:}\\ \nu ‘=\left(\frac{v+{\text{v}}_{\text{e}}}{v}\right)\nu =\left(\frac{1450+100}{1450}\right)\times 40=42.76kHz\\ \text{The frequency }\left(\nu ‘‘\right)\text{ received by the enemy submarine}\text{is given as:}\\ \nu ‘‘=\left(\frac{v}{v+{\text{v}}_s}\right)\nu ‘\\ \text{Here,}{\text{v}}_{\text{s}}=100{\text{ms}}^{-1}\\ \therefore \nu ‘‘=\left(\frac{1450}{1450+100}\right)\times 42.76=45.93\text{kHz}\end{array}

Q24. Read the following two statements below carefully and state, with reasons, if it is true or false.</>

(a) The Young’s modulus of rubber is greater than that of steel;

(b) The stretching of a coil is determined by its shear modulus.

Ans:

\begin{array}{}\begin{array}{}\text{(a) The}\text{given}\text{statement}\text{is}\text{false,}\text{because}\text{for a given }\\ \text{stress}\text{there}\text{is}\text{more}\text{strain in rubber than steel}\text{.}\\ \text{Young’s modulus}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \therefore \text{For a constant}\text{value}\text{of}\text{stress:}\text{Y}\propto \frac{\text{1}}{\text{Strain}}\\ \therefore \text{The}\text{Young’s}\text{modulus}\text{of rubber is less as}\text{compared}\text{to}\\ \text{it}\text{is}\text{for steel}\text{.}\end{array}\\ \text{(b) Shear modulus is given by the ratio of the applied stress to }\\ \text{the change in the shape of a body}\text{. The stretching of a coil changes}\\ \text{ its shape without any change in the length of the wire used in the }\\ \text{coil}\text{. Therefore, shear modulus of elasticity is involved in it}\text{.}\end{array}

Q25. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

Ans:

\begin{array}{l}\text{Here,}\text{Temperature of helium atom,}{\text{T}}_{\text{He}}=–\text{2}0°\text{C}\\ =\text{253 K}\\ \text{Atomic mass of the}\text{argon}\text{gas,}{\text{M}}_{\text{Ar}}=\text{39}.\text{9 u}\\ \text{Atomic mass of the}\text{helium}\text{gas},{\text{M}}_{\text{He}}=\text{4}.0\text{ u}\\ \text{Let}\text{rms speed of argon}\text{gas}\text{=}{\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}\\ \text{Let rms speed of helium}\text{gas}\text{=}{\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\\ \text{Rms speed of the}\text{argon gas}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}=\sqrt{\frac{3R{T}_{Ar}}{M_{Ar}}}\to \left(\text{i}\right)\\ \text{Where,}\text{R}\text{=}\text{Universal gas constant}\\ {\text{T}}_{\text{Ar}}\text{=}\text{Temperature of argon gas}\\ \text{Rms speed of the}\text{helium gas}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}\to \left(\text{ii}\right)\\ \text{As}\text{per}\text{the}\text{question:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}={\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\\ \sqrt{\frac{3R{T}_{Ar}}{M_{Ar}}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}\\ \frac{T_{Ar}}{M_{Ar}}=\frac{T_{He}}{M_{He}}\\ T_{Ar}=\frac{T_{He}}{M_{He}}\times M_{Ar}=\frac{253}{4}\times 39.9\\ =\text{2523}.\text{675}=\text{2}.\text{52}\times \text{1}{0}^{\text{3}}\text{K}\\ \therefore \text{Temperature of the argon gas}\text{atom}\text{=}\text{2}.\text{52}\times \text{1}{0}^{\text{3}}\text{ K}\end{array}

Q26. Which of the following is the most precise device for measuring length :
(a) a vernier calipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?

Ans:

\begin{array}{}\text{(a)}\text{Least}\text{count}\text{of}\text{the}\text{given}\text{vernier}\text{callipers}\\ \text{= }\frac{\text{Value}\text{of}\text{one}\text{main}\text{scale}\text{division}}{\text{Total}\text{number}\text{of}\text{divisions}\text{on}\text{the}\text{sliding}\text{scale}}\\ \text{= }\frac{\text{1}\text{mm}}{\text{20}}\text{ = 0}\text{.05}\text{mm = 0}\text{.005}c\text{m}\\ \text{(b)}\text{Least}\text{count}\text{of}\text{screw}\text{gauge}\\ \text{= }\frac{\text{Pitch}\text{of}\text{screw}\text{gauge}}{\text{Total}\text{no}\text{.}\text{of}\text{divisions}\text{on}\text{circular}\text{scale}}\\ \text{= }\frac{\text{1}}{\text{100}}\text{ mm = 0}\text{.01 mm = 0}\text{.001}\text{cm}\\ \text{(c)}\text{Here,}\text{wavelength}\text{of}\text{light,}\text{λ}\approx {\text{10}}^{\text{-5}}\text{ cm = 0}\text{.00001}\text{cm}\\ \text{The}\text{most}\text{precise}\text{device}\text{is}\text{that}\text{whose}\text{least}\text{count}\text{is}\text{minimum}\text{.}\\ \therefore \text{An}\text{optical}\text{instrument}\text{is}\text{the}\text{most}\text{precise}\text{device}\text{.}\end{array}

Q27. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev s^-1 at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.

Ans:

\begin{array}{}\text{Here,}\text{mass, m = 14}\text{.5 kg}\\ \text{Length of steel wire, l = 1}\text{.0 m}\\ {\text{Angular velocity, ω = 2 revs}}^{\text{-1}}\\ \text{Area of Cross-section}\text{of the wire, a = 0}{\text{.065 cm}}^{\text{2}}\text{ = 0}{\text{.065 ×10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{When the mass is at the lowest point of its path:}\\ \text{Let}\text{elongation of the wire = ΔL }\\ \text{Total force exerted}\text{on the}\text{mass,}\text{when it is placed at }\\ \text{the lowest}\text{position of the vertical circleis}\text{given}\text{as:}\\ {\text{F = mg + mlω}}^{\text{2}}\\ \text{F =14}\text{.5 kg × 9}{\text{.8 ms}}^{-2}\text{ +14}\text{.5 kg ×1 m×}{\left({\text{2 revs}}^{\text{-1}}\right)}^{\text{2}}\\ \text{F = 200}\text{.1 N}\\ \text{The}\text{relation}\text{for}\text{Young’s modulus is}\text{given}\text{as,}\\ \text{Y =}\frac{\text{Stress}}{\text{Strain}}\text{ = }\frac{\frac{\text{F}}{\text{A}}}{\frac{\text{Δl}}{\text{l}}}\text{=}\frac{\text{Fl}}{\text{AΔl}}\\ \therefore \text{Δl = }\frac{\text{Fl}}{\text{AY}}\text{(i)}\\ \text{Young’s modulus}\text{for}\text{Steel =}{\text{2 × 10}}^{\text{11}}\text{Pa}\\ \text{Applying}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \text{Δl = }\frac{\text{200}\text{.1 N ×1 m}}{\text{0}{\text{.065×10}}^{\text{-4}}{\text{×2×10}}^{\text{11}}\text{ Pa}}\\ \text{Δl = 1}{\text{.539× 10}}^{\text{–4}}\text{m}\\ \therefore \text{Elongation of the wire = 1}{\text{.539 × 10}}^{\text{–4}}\text{m }\end{array}

Q28. There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075N/m and density 10³kg/m³. The bubble is at a depth of 10.0 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure?

OR

A person weighing 60 kg takes in 2000 kcal diet in a day. If this energy were to be used in heating the person without any losses, what would be his rise in temperature? Given specific heat of human body is 0.83 cal g^{-1 0}C^-1 .   

Ans:

Here  r = 1.0mm = 10^-3m;S = 0.075 N/m

OR

Q29. Two bodies of masses 5kg and 3kg moving in the same direction along the same straight line with velocities 5m/s and 3m/s respectively suffers one dimensional elastic collision. Find the velocities of two bodies after the collision.

OR

A sphere of mass 2 kg is attached to a string 4 m long and is whirled in a horizontal circle. The string can withstand a maximum tension of 32 N. Calculate maximum velocity of revolution that can be given to the stone without breaking the string.

 

Ans:

 

OR

T= 32N; m = 2kg; r = 4m

The tension in the string provides the necessary centripetal acceleration.
So, T = m r

$\omega^2$

ω2​

$\omega^2$

ω2​=  T / mr

= 32N / (2kg x 4m)

$\omega$

ω​= 2 rad /s

So, the velocity of revolution, v =

$\omega$

ω​ x r

= 2 rad/s x 4m

= 8 m/s.

 

Q30. A block of mass 2 kg hanged with a rope is going upward with an acceleration of 4m/s², calculate tension.

Ans:”

The equation of motion of block is given by

F   = T – mg

Ma = T – mg

T= mg + ma

= m (g + a)

= 2 (9.8 + 2)

T= 2 x 11.8 = 23.6N

Q31. Two masses m₁ and m₂ are separated by distance r. Find the moment of inertia of this arrangement about an axis passing through the centre of mass and perpendicular to the line joining them.

OR

A solid sphere rolls up an inclined plane of angle of inclination 30⁰.At the bottom of the inclined plane the centre of mass of the sphere has speed of 5 m/sec.Find how far up the sphere will go?

Ans: If centre of mass is origin,

 

 

 

OR

Let a be the deacceleration of the sphere

Q32. Two wires A and B of lengths 2m and 1.4m respectively have radii of 0.45 × 10^-2 m & 0.5 × 10^–2 m. Weights of 7 kg and 12 kg are attached to the wires as per the arrangement shown below.
Ans:

Q33. The velocity of a small ball of mass 10 g and density 7.8g/cc becomes constant when dropped in a container filled with glycerine. If the density of glycerine is 1.3g/cc ,find the viscous force acting on the ball?

OR

Water is flowing in a river. If the velocity of a layer at a distance 10cm from the bottom is 20cm/sec ,find the velocity of layer at a height of 40cm from the bottom?

Ans:

Here mass of the ball is 10g

We know,

Q34. The cylindrical tube of a spray pump has a cross section of 8.0cm² ,one of which has 40 fine holes each of diameter 1.0mm. If the liquid flows inside the tube at the rate of 1.5m per minute, what is the speed of ejection of the liquid through the holes?

Ans:

Area of cross section of tube, a₁= 8.0cm² = 8.0X10^-4m²

No of holes = 40
Diameter of each hole D = 1mm = 10^-3m

Q35. A stationary source emits a sound towards a wall moving towards it with a velocity u. Speed of sound in air is v. Find the fractional change in wavelength of the sound sent and the sound reflected which is observed at source.

Ans:

 

Wavelength sent = λ. Let actual frequency = f

Here wall is moving towards source. So frequency observed at wall