Question 1

Question: Which types of motion does a coin undergo as it moves down an inclined plane?

• (a) rotational and translational

• (b) circular and random

• (c) translational

• (d) rotational

Topic: Types of Motion
Correct Option: (a)
Detailed Explanation:
When a coin rolls down an inclined plane, it exhibits two simultaneous motions:

1. Translational Motion: The center of mass of the coin moves in a straight line down the slope.

2. Rotational Motion: The coin spins around its own center/axis as it moves.
   Therefore, the combined motion is rotational and translational.

Question 2

Question: An object of mass 20kg is released from a height of 5m above the ground level. The kinetic energy of the object just before hitting the ground (a)

• (a) 1000J

• (b) 2000J

• (c) 1500J

• (d) 4000J

Topic: Work, Energy, and Power (Conservation of Energy)
Correct Option: (a)
Detailed Explanation:
According to the law of conservation of energy, the Potential Energy (P.E.) at the top equals the Kinetic Energy (K.E.) just before hitting the ground (assuming negligible air resistance).

$$P.E.=mgh$$

m=20kg,g=10m/s2,h=5mm = 20\text{kg}, \quad g = 10\text{m/s}^2, \quad h = 5\text{m}m=20kg,g=10m/s2,h=5m

$$P.E.=20\times 10\times 5=1000\text{J}$$

Therefore, K.E. at the bottom = 1000J.

Question 3

Question: The small droplets of water that form on the grass in the early hours of the morning is

• (a) dew

• (b) fog

• (c) hail

• (d) mist

Topic: Change of State / Meteorology
Correct Option: (a)
Detailed Explanation:

• Dew occurs when atmospheric vapor condenses directly onto cool surfaces (like grass) at night.

• Fog and Mist are water droplets suspended in the air.

• Hail is frozen precipitation.

Question 4

Question: The equation

PaVbTc=constantP^a V^b T^c = \text{constant}PaVbTc=constant

reduces to Boyle’s law if

• (a) a = 0, b = 1, c = -1

• (b) a = 1, b = 1, c = 0

• (c) a = 1, b = 1, c = 1

• (d) a = 1, b = 1, c = -1

Topic: Gas Laws
Correct Option: (b)
Detailed Explanation:
Boyle’s Law states that for a fixed mass of gas at constant temperature, Pressure is inversely proportional to Volume. The formula is

$PV=\text{constant}$

.
Looking at

PaVbTcP^a V^b T^cPaVbTc

:

• We need

  P1P^1P1

  and

  V1V^1V1

  (so

  $a=1,b=1$

  ).

   

• Temperature (

  $T$

  ) does not appear in the Boyle’s law equation (it is constant), which mathematically is equivalent to

  T0T^0T0

  (since anything to the power of 0 is 1). So

  $c=0$

  .
  Matches option (b).

   

Question 5

Question: The distance between two successive troughs of a wave is 30cm and the velocity is 300ms⁻¹. Calculate the frequency.

• (a) 100Hz

• (b) 2250Hz

• (c) 1000Hz

• (d) 225Hz

Topic: Waves
Correct Option: (c)
Detailed Explanation:

• Distance between successive troughs = Wavelength (

  $\lambda$

  ) = 30cm = 0.3m.

   

• Velocity (

  $v$

  ) = 300 m/s.

   

• Formula:

  $v=f\lambda$

   

•  

  f=vλ=3000.3=30003=1000Hzf = \frac{v}{\lambda} = \frac{300}{0.3} = \frac{3000}{3} = 1000\text{Hz}f=λv​=0.3300​=33000​=1000Hz

  .

   

Question 6

Question: In a compound microscope, the objective and the eyepiece focal lengths are

• (a) at infinity

• (b) long

• (c) the same

• (d) short

Topic: Optical Instruments
Correct Option: (d)
Detailed Explanation:
To achieve high magnification, both lenses in a compound microscope must have short focal lengths.

• Magnifying power is inversely proportional to focal length.

• The objective lens has a very short focal length (to form a real, magnified image).

• The eyepiece has a short focal length (to act as a magnifier for the first image).

Question 7

Question: What is the best method of demagnetizing a steel bar magnet?

• (a) solenoid method

• (b) hammering it

• (c) heating it

• (d) rough handling it

Topic: Magnetism
Correct Option: (a)
Detailed Explanation:
While heating and hammering can demagnetize a magnet, the electrical method (solenoid) is the most effective and controlled “scientific” method. Placing the magnet in a solenoid carrying an Alternating Current (AC) and slowly withdrawing it in an East-West direction destroys the magnetic domains completely.

Question 8

Question: The magnitude of the angle of dip at the equator is

• (a) 180⁰

• (b) 0⁰

• (c) 90⁰

• (d) 60⁰

Topic: Earth’s Magnetism
Correct Option: (b)
Detailed Explanation:

• At the magnetic Equator, the earth’s magnetic field lines run parallel to the horizontal. Thus, the angle of dip is 0⁰.

• At the magnetic Poles, the field lines are vertical, so the dip is 90⁰.

Question 9

Question: A glass bottle of initial volume

2×104cm32 \times 10^4 \text{cm}^32×104cm3

is heated from 20°C to 50°C. If the linear expansivity of glass is

9×10−6°C−19 \times 10^{-6} \text{°C}^{-1}9×10−6°C−1

, the volume of the bottle at 50°C is

• (a) 20013.5cm³

• (b) 20016.2cm³

• (c) 2005.4cm³

• (d) 20008.1cm³

Topic: Thermal Expansion
Correct Option: (b)
Detailed Explanation:

1. Linear expansivity (

   $\alpha$

   ) =

   9×10−6K−19 \times 10^{-6} \text{K}^{-1}9×10−6K−1

   .

    

2. Cubic (Volume) expansivity (

   $\gamma$

   ) =

   3α=3×9×10−6=27×10−6K−13\alpha = 3 \times 9 \times 10^{-6} = 27 \times 10^{-6} \text{K}^{-1}3α=3×9×10−6=27×10−6K−1

   .

    

3. Change in temp (

   $\Delta \theta$

   ) =

   $50-20=30\text{°C}$

   .

    

4. Original Volume (

   V1V_1V1​

   ) =

   20,000cm320,000 \text{cm}^320,000cm3

   .

    

5. Change in Volume (

   $\Delta V$

   ) =

   V1γΔθV_1 \gamma \Delta \thetaV1​γΔθ

   
   

   ΔV=20,000×(27×10−6)×30\Delta V = 20,000 \times (27 \times 10^{-6}) \times 30ΔV=20,000×(27×10−6)×30

   
   

   ΔV=2×104×27×10−6×3×101\Delta V = 2 \times 10^4 \times 27 \times 10^{-6} \times 3 \times 10^1ΔV=2×104×27×10−6×3×101

   
   

   ΔV=162×10−1=16.2cm3\Delta V = 162 \times 10^{-1} = 16.2 \text{cm}^3ΔV=162×10−1=16.2cm3

    

6. New Volume =

   20,000+16.2=20016.2 cm320,000 + 16.2 = \textbf{20016.2 cm}^320,000+16.2=20016.2 cm3

   .

    

Question 10

Question: When the linear momentum of a body is constant, the net force acting on it (a) is

• (a) zero

• (b) decreases

• (c) increases

• (d) remains constant

Topic: Newton’s Second Law
Correct Option: (a)
Detailed Explanation:
Newton’s Second Law states that Force is the rate of change of momentum (

F=ΔptF = \frac{\Delta p}{t}F=tΔp​

).
If momentum (

$p$

) is constant, the change in momentum (

$\Delta p$

) is zero.
Therefore, Force = 0.

Question 11

Question: A device that converts mechanical energy into electrical energy is (a)

• (a) transformer

• (b) dynamo

• (c) an electric motor

• (d) an induction coil

Topic: Electromagnetic Induction
Correct Option: (b)
Detailed Explanation:

• A Dynamo (or Generator) converts Mechanical Energy

  $\to$

  Electrical Energy.

   

• An Electric Motor converts Electrical

  $\to$

  Mechanical.

   

Question 12

Question: A quantity of water at 20°C is mixed with another quantity of water at 70°C. The final steady temperature of the mixture is 40°C. Determine the ratio of the mass of the hot water to that of the cold water.

• (a) 2:1

• (b) 2:3

• (c) 5:2

• (d) 2:7

Topic: Heat Energy (Method of Mixtures)
Correct Option: (b)
Detailed Explanation:
Heat lost by hot water = Heat gained by cold water.
Let

mhm_hmh​

be mass of hot water and

mcm_cmc​

be mass of cold water. Specific heat capacity

$c$

cancels out.

mhc(70−40)=mcc(40−20)m_h c (70 - 40) = m_c c (40 - 20)mh​c(70−40)=mc​c(40−20)

mh(30)=mc(20)m_h (30) = m_c (20)mh​(30)=mc​(20)

mhmc=2030=23\frac{m_h}{m_c} = \frac{20}{30} = \frac{2}{3}mc​mh​​=3020​=32​

Ratio is 2:3.

Question 13

Question: The quantity of a note depends on its

• (a) pitch

• (b) frequency

• (c) amplitude

• (d) overtone

Topic: Sound Waves
Correct Option: (d)
Detailed Explanation:

• Quality (or Timbre) allows us to distinguish between different instruments playing the same note. It depends on the number and intensity of overtones (harmonics).

• Pitch depends on frequency.

• Loudness depends on amplitude.

Question 14

Question: An ammeter can be adapted to measure potential difference by using a (a)

• (a) rheostat

• (b) multiplier

• (c) shunt

• (d) resistance box

Topic: Electrical Instrumentation
Correct Option: (b)
Detailed Explanation:
To measure potential difference (act as a voltmeter), a galvanometer/ammeter needs a high resistance connected in series. This high resistance is called a multiplier.
(Note: A shunt is a low resistance in parallel used to convert a galvanometer to an ammeter).

Question 15

Question: A transformer has 800 turns of wire in the primary coil and 80 turns in the secondary coil. If the input voltage is 150V, calculate the magnitude of the output voltage.

• (a) 15V

• (b) 150V

• (c) 36V

• (d) 55V

Topic: Transformers
Correct Option: (a)
Detailed Explanation:
Formula:

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}Vp​Vs​​=Np​Ns​​

Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p}Vs​=Vp​×Np​Ns​​

Vs=150×80800V_s = 150 \times \frac{80}{800}Vs​=150×80080​

Vs=150×110=15VV_s = 150 \times \frac{1}{10} = 15\text{V}Vs​=150×101​=15V

.

Question 16

Question: An element of nucleon number P and atomic number Q emits an alpha particle from its nucleus. The resultant numbers of the new element formed are respectively

• (a) P+2 and Q-2

• (b) P-4 and Q+2

• (c) P-4 and Q-2

• (d) P+2 and Q+2

Topic: Radioactivity
Correct Option: (c)
Detailed Explanation:
An alpha particle (

$\alpha$

) is a Helium nucleus (

24He^4_2\text{He}24​He

). It has a mass number of 4 and an atomic number of 2.
When emitted:

• Nucleon number (Mass) decreases by 4 (

  $P\to P-4$

  ).

   

• Atomic number (Proton) decreases by 2 (

  $Q\to Q-2$

  ).

   

Question 17

Question: Calculate the total distance covered by a train before coming to rest if its initial speed is

30ms−130\text{ms}^{-1}30ms−1

with a constant retardation of

0.2ms−20.2\text{ms}^{-2}0.2ms−2

.

• (a) 1500m

• (b) 2750m

• (c) 2100m

• (d) 2250m

Topic: Equations of Motion
Correct Option: (d)
Detailed Explanation:
Use the equation:

v2=u2+2asv^2 = u^2 + 2asv2=u2+2as

• Final velocity

  $v=0$

  (comes to rest)

   

• Initial velocity

  $u=30$

   

• Acceleration

  $a=-0.2$

  (retardation)
  

  0=302+2(−0.2)s0 = 30^2 + 2(-0.2)s0=302+2(−0.2)s

  
  

  $0=900-0.4s$

  
  

  $0.4s=900$

  
  

  s=9000.4=90004=2250ms = \frac{900}{0.4} = \frac{9000}{4} = 2250\text{m}s=0.4900​=49000​=2250m

  .

   

Question 18

Question: An object of volume

1m31\text{m}^31m3

and mass 2kg is fully immersed in a liquid of density

1kgm−31\text{kgm}^{-3}1kgm−3

. Calculate its apparent weight.

• (a) 1N

• (b) 2N

• (c) 10N

• (d) 20N

Topic: Archimedes’ Principle
Correct Option: (c)
Detailed Explanation:

1. Real Weight (

           $W$ 

   ):

   $mg=2\times 10=20\text{N}$

   .

    

2. Upthrust (

           $U$ 

   ): Weight of liquid displaced =

   Volume×ρliquid×g\text{Volume} \times \rho_{\text{liquid}} \times gVolume×ρliquid​×g

   
   

   $U=1\times 1\times 10=10\text{N}$

   .

    

3. Apparent Weight:

   $W-U=20-10=10\text{N}$

   .

    

Question 19

Question: Which of the following is not a factor that can increase the rate of evaporation of water in a lake?

• (a) increase in the kinetic energy of the molecules of water

• (b) rise in temperature

• (c) increase in the pressure of the atmosphere

• (d) increase in the average speed of the molecules of water

Topic: Heat (Evaporation)
Correct Option: (c)
Detailed Explanation:
Evaporation is the escape of energetic molecules from the surface.

• High Temp, High KE, and Wind speed increase evaporation.

• High Atmospheric Pressure pushes down on the liquid surface, making it harder for molecules to escape, thus decreasing the rate.

Question 20

Question: Where can a man place his face to obtain an enlarged image when using a concave mirror to shave?

• (a) at infinity

• (b) at the principal focus

• (c) between the centre of curvature and the principal focus

• (d) between the principal focus and the pole

Topic: Optics (Mirrors)
Correct Option: (d)
Detailed Explanation:
Concave mirrors produce a Virtual, Erect, and Enlarged image only when the object is placed between the Focus (F) and the Pole (P). This is the principle used in shaving mirrors.

Question 21

Question: A particle carrying a charge of

1×10−8C1 \times 10^{-8}\text{C}1×10−8C

enters a magnetic field… at right angles to the field. If the force on this particle is

1.8×10−8N1.8 \times 10^{-8}\text{N}1.8×10−8N

, what is the magnitude of the field? (Velocity is

3×102ms−13 \times 10^2 \text{ms}^{-1}3×102ms−1

).

• (a)

  6.0×10−1T6.0 \times 10^{-1}\text{T}6.0×10−1T

   

• (b)

  6×103T6 \times 10^3\text{T}6×103T

   

• (c)

          6×10−3T6 \times 10^{-3}\text{T}6×10−3T 

   

• (d)

  6×104T6 \times 10^4\text{T}6×104T

   

Topic: Electromagnetism
Correct Option: (c)
Detailed Explanation:
Formula:

$F=qvB\sin \theta$

(where

θ=90∘\theta = 90^\circθ=90∘

, so

$\sin 90=1$

)

B=FqvB = \frac{F}{qv}B=qvF​

B=1.8×10−8(1×10−8)×(3×102)B = \frac{1.8 \times 10^{-8}}{(1 \times 10^{-8}) \times (3 \times 10^2)}B=(1×10−8)×(3×102)1.8×10−8​

Powers of

10−810^{-8}10−8

cancel out.

B=1.8300=183000=61000=6×10−3TB = \frac{1.8}{300} = \frac{18}{3000} = \frac{6}{1000} = 6 \times 10^{-3}\text{T}B=3001.8​=300018​=10006​=6×10−3T

.

Question 22

Question: A glass block of thickness 10cm and refractive index 3/2 is placed on an object. If an observer views the object vertically, the displacement of the object is

• (a) 8.5cm

• (b) 3.33cm

• (c) 5.00cm

• (d) 6.67cm

Topic: Refraction
Correct Option: (b)
Detailed Explanation:
Vertical displacement (Lateral displacement)

d=t(1−1n)d = t (1 - \frac{1}{n})d=t(1−n1​)

d=10(1−11.5)=10(1−23)d = 10 (1 - \frac{1}{1.5}) = 10 (1 - \frac{2}{3})d=10(1−1.51​)=10(1−32​)

d=10(13)=3.33cmd = 10 (\frac{1}{3}) = 3.33\text{cm}d=10(31​)=3.33cm

.

Question 23

Question: A girl stands on a scale in a lift. If the reading on the scale is less than her weight, then the lift is moving

• (a) downwards with uniform acceleration

• (b) downwards with uniform speed

• (c) upwards with uniform speed

• (d) upwards with uniform acceleration

Topic: Mechanics (Lift Motion)
Correct Option: (a)
Detailed Explanation:

• Scale reading (

  $R$

  ) is the Normal Reaction.

   

• When accelerating downwards, the net force is down (

  $mg-R=ma$

  ).

   

• So,

  $R=m(g-a)$

  .

   

• Since

  $R<mg$

  (reading is less than weight), the lift must be accelerating downwards.

   

Question 24

Question: In the formation of sea breeze, wind blows from (a)

• (a) sea to land

• (b) land to sea

• (c) sky to land

• (d) sea to sky

Topic: Heat Transfer (Convection)
Correct Option: (a)
Detailed Explanation:
Sea breeze occurs during the day. The land heats up faster than the sea. Hot air over the land rises (low pressure), and cool air from the sea blows towards the land to replace it. Hence, Sea to Land.

Question 25

Question: A simple pendulum of length 0.4m has a period of 2s. What is the period of a similar pendulum of length 0.8m at the same place?

• (a)

  2s\sqrt{2}\text{s}2​s

   

• (b)

          22s2\sqrt{2}\text{s}22​s 

   

• (c)

  $4\text{s}$

   

• (d)

  $8\text{s}$

   

Topic: Simple Harmonic Motion
Correct Option: (b)
Detailed Explanation:
Period

T∝LT \propto \sqrt{L}T∝L​

.

T2T1=L2L1\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}T1​T2​​=L1​L2​​​

T22=0.80.4=2\frac{T_2}{2} = \sqrt{\frac{0.8}{0.4}} = \sqrt{2}2T2​​=0.40.8​​=2​

T2=22sT_2 = 2\sqrt{2}\text{s}T2​=22​s

.

Question 26

Question: An object is moving with a velocity of

10ms−110\text{ms}^{-1}10ms−1

. At what height must a similar body be situated to have a potential energy equal in value to the kinetic energy of the moving body?

• (a) 5m

• (b) 2.5m

• (c) 10m

• (d) 3m

Topic: Mechanical Energy
Correct Option: (a)
Detailed Explanation:
Set

$P.E.=K.E.$

mgh=12mv2mgh = \frac{1}{2}mv^2mgh=21​mv2

gh=v22gh = \frac{v^2}{2}gh=2v2​

10×h=102210 \times h = \frac{10^2}{2}10×h=2102​

$10h=50⟹h=5\text{m}$

.

Question 27

Question: A wave that travels through stretched strings is known as (a)

• (a) microwave

• (b) seismic wave

• (c) mechanical wave

• (d) electromagnetic wave

Topic: Waves
Correct Option: (c)
Detailed Explanation:
Waves on strings require a material medium (the string) to travel. Therefore, they are Mechanical waves. Specifically, they are transverse mechanical waves.

Question 28

Question: What is the velocity of sound at 100°C if the velocity of sound at 0°C is

340ms−1340\text{ms}^{-1}340ms−1

?

• (a)

  440ms−1440\text{ms}^{-1}440ms−1

   

• (b)

          397ms−1397\text{ms}^{-1}397ms−1 

   

• (c)

  497ms−1497\text{ms}^{-1}497ms−1

   

• (d)

  240ms−1240\text{ms}^{-1}240ms−1

   

Topic: Sound Waves
Correct Option: (b)
Detailed Explanation:
Velocity of sound is proportional to the square root of absolute temperature (Kelvin).

v∝Tv \propto \sqrt{T}v∝T​

v2v1=T2T1\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}v1​v2​​=T1​T2​​​

T1=0+273=273KT_1 = 0 + 273 = 273\text{K}T1​=0+273=273K

.

T2=100+273=373KT_2 = 100 + 273 = 373\text{K}T2​=100+273=373K

.

v2=340×373273≈340×1.366≈340×1.169≈397ms−1v_2 = 340 \times \sqrt{\frac{373}{273}} \approx 340 \times \sqrt{1.366} \approx 340 \times 1.169 \approx 397\text{ms}^{-1}v2​=340×273373​​≈340×1.366​≈340×1.169≈397ms−1

.

Question 29

Question: The angle of deviation of light of various colours passing through a triangular prism increases in the order

• (a) red

          $\to$ 

  green

  $\to$

  blue

   

• (b) blue

  $\to$

  green

  $\to$

  red

   

• (c) green

  $\to$

  violet

  $\to$

  blue

   

• (d) blue

  $\to$

  red

  $\to$

  green

   

Topic: Dispersion of Light
Correct Option: (a)
Detailed Explanation:
Red light has the longest wavelength and deviates (bends) the least. Violet/Blue light has the shortest wavelength and deviates the most.
Order of increasing deviation: Red < Orange < Yellow < Green < Blue < Indigo < Violet.
Option (a) follows this order correctly.

Question 30

Question: Capacitors are used in the induction coil to (a)

• (a) dissipate energy

• (b) control circuits

• (c) prevent electric sparks

• (d) distort the electric fields

Topic: Electronics
Correct Option: (c)
Detailed Explanation:
In an induction coil, a capacitor is connected across the “make-and-break” contact points. Its function is to absorb the induced current when the circuit breaks, thereby preventing sparking at the contact points and ensuring a rapid collapse of the magnetic field.

Question 31

Question: If a current of 2.5A flows through electrolyte for 3 hours and 1.8g of a substance [is deposited]… what mass will be deposited if a current of 4A flows through it for 4.8 hours?

• (a) 4.8g

• (b) 3.2g

• (c) 2.4g

• (d) 4.6g

Topic: Electrolysis (Faraday’s First Law)
Correct Option: (d)
Detailed Explanation:

$m=ZIt$

.
Since it is the same substance, Z is constant.

m2m1=I2t2I1t1\frac{m_2}{m_1} = \frac{I_2 t_2}{I_1 t_1}m1​m2​​=I1​t1​I2​t2​​

m2=1.8×4×4.82.5×3=1.8×19.27.5m_2 = 1.8 \times \frac{4 \times 4.8}{2.5 \times 3} = 1.8 \times \frac{19.2}{7.5}m2​=1.8×2.5×34×4.8​=1.8×7.519.2​

m2=1.8×2.56=4.608gm_2 = 1.8 \times 2.56 = 4.608\text{g}m2​=1.8×2.56=4.608g

.
Approx 4.6g.

Question 32

Question: Which of the following units is equivalent to watt?

• (a)

  kgm2s−1kg m^2 s^{-1}kgm2s−1

   

• (b)

  kgms−2kg m s^{-2}kgms−2

   

• (c)

  kgms−2kg m s^{-2}kgms−2

   

• (d)

          kgm2s−3kg m^2 s^{-3}kgm2s−3 

   

Topic: Units and Dimensions
Correct Option: (d)
Detailed Explanation:
Watt is Power.
Power =

WorkTime=Force×DistanceTime\frac{\text{Work}}{\text{Time}} = \frac{\text{Force} \times \text{Distance}}{\text{Time}}TimeWork​=TimeForce×Distance​

Force = Mass

$\times$

Acc =

kg⋅m/s2kg \cdot m/s^2kg⋅m/s2

Power =

(kg⋅m/s2)⋅ms=kg⋅m2⋅s−3\frac{(kg \cdot m/s^2) \cdot m}{s} = kg \cdot m^2 \cdot s^{-3}s(kg⋅m/s2)⋅m​=kg⋅m2⋅s−3

.

Question 33

Question: A diverging lens of focal length 30cm produces an image 20cm from the lens. Determine the object distance.

• (a) 10cm

• (b) 12cm

• (c) 60cm

• (d) 50cm

Topic: Light (Lenses)
Correct Option: (c)
Detailed Explanation:
Lens formula:

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}f1​=v1​+u1​

Diverging lens:

$f$

is negative (

$-30$

). Image is always virtual, so

$v$

is negative (

$-20$

).

1−30=1−20+1u\frac{1}{-30} = \frac{1}{-20} + \frac{1}{u}−301​=−201​+u1​

1u=120−130=3−260=160\frac{1}{u} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}u1​=201​−301​=603−2​=601​

$u=60\text{cm}$

.

Question 34

Question: A string under tension produces a note of frequency 14Hz. Determine the frequency when the tension is quadrupled.

• (a) 56Hz

• (b) 28Hz

• (c) 18Hz

• (d) 14Hz

Topic: Waves in Strings
Correct Option: (b)
Detailed Explanation:
Frequency

f∝Tf \propto \sqrt{T}f∝T​

.
If Tension

$T$

becomes

$4T$

, the new frequency

f2∝4T=2Tf_2 \propto \sqrt{4T} = 2\sqrt{T}f2​∝4T​=2T​

.
The frequency doubles.

$14\times 2=28\text{Hz}$

.

Question 35

Question: The work function of a metal is

2.56×10−19J2.56 \times 10^{-19}\text{J}2.56×10−19J

. Calculate the frequency of a photon whose energy is required to eject from the metal an electron with kinetic energy of 3.0eV.

• (a)

          1.12×1015Hz1.12 \times 10^{15}\text{Hz}1.12×1015Hz 

   

• (b)

  2.2×1014Hz2.2 \times 10^{14}\text{Hz}2.2×1014Hz

   

• (c)

  1.12×105Hz1.12 \times 10^{5}\text{Hz}1.12×105Hz

   

• (d)

  4.4×1014Hz4.4 \times 10^{14}\text{Hz}4.4×1014Hz

   

Topic: Photoelectric Effect
Correct Option: (a)
Detailed Explanation:

$E=hf=W+K.E.$

W=2.56×10−19JW = 2.56 \times 10^{-19}\text{J}W=2.56×10−19J

.

K.E.=3.0eV=3.0×1.6×10−19J=4.8×10−19JK.E. = 3.0\text{eV} = 3.0 \times 1.6 \times 10^{-19}\text{J} = 4.8 \times 10^{-19}\text{J}K.E.=3.0eV=3.0×1.6×10−19J=4.8×10−19J

.
Total Energy

E=(2.56+4.8)×10−19=7.36×10−19JE = (2.56 + 4.8) \times 10^{-19} = 7.36 \times 10^{-19}\text{J}E=(2.56+4.8)×10−19=7.36×10−19J

.

f=Eh=7.36×10−196.6×10−34≈1.115×1015Hzf = \frac{E}{h} = \frac{7.36 \times 10^{-19}}{6.6 \times 10^{-34}} \approx 1.115 \times 10^{15}\text{Hz}f=hE​=6.6×10−347.36×10−19​≈1.115×1015Hz

.

Question 36

Question: When a red rose flower is observed in blue light, what colour does the observer see?

• (a) blue

• (b) red

• (c) black

• (d) yellow

Topic: Colour of Objects
Correct Option: (c)
Detailed Explanation:
A red rose appears red because it reflects red light and absorbs all other colors. When illuminated by blue light, the rose absorbs the blue light. Since there is no red light to reflect, no light reaches the observer’s eye, and it appears black.

Question 37

Question: The density

$\rho$

of a spherical ball of diameter

$d$

and mass

$m$

is given by (a)

$\rho =\dots$

• (a)

  $\dots$

   

• (b)

  $\dots$

   

• (c)

  $\dots$

   

• (d)

          ρ=6mπd3\rho = \frac{6m}{\pi d^3}ρ=πd36m​ 

   

Topic: Density
Correct Option: (d)
Detailed Explanation:
Density

ρ=mV\rho = \frac{m}{V}ρ=Vm​

.
Volume of sphere

V=43πr3V = \frac{4}{3} \pi r^3V=34​πr3

.
Since

r=d2r = \frac{d}{2}r=2d​

,

V=43π(d2)3=43πd38=πd36V = \frac{4}{3} \pi (\frac{d}{2})^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{6}V=34​π(2d​)3=34​π8d3​=6πd3​

.

ρ=mπd36=6mπd3\rho = \frac{m}{\frac{\pi d^3}{6}} = \frac{6m}{\pi d^3}ρ=6πd3​m​=πd36m​

.

Question 38

Question: A coin is pushed from the edge of a laboratory bench with a horizontal velocity of

15ms−115\text{ms}^{-1}15ms−1

. If the height of the bench from the floor is 1.5m, calculate the distance from the foot of the bench to the point of impact with the floor.

• (a) 8.22m

• (b) 2.25m

• (c) 1.5m

• (d) 1.41m

Topic: Projectile Motion
Correct Option: (a)
Detailed Explanation:

1. Find time of flight (

   $t$

   ):

   h=12gt2h = \frac{1}{2}gt^2h=21​gt2

   
   

   1.5=0.5(10)t2  ⟹  1.5=5t21.5 = 0.5(10)t^2 \implies 1.5 = 5t^21.5=0.5(10)t2⟹1.5=5t2

   
   

   t2=0.3  ⟹  t=0.3≈0.5477st^2 = 0.3 \implies t = \sqrt{0.3} \approx 0.5477\text{s}t2=0.3⟹t=0.3​≈0.5477s

   .

    

2. Horizontal Distance (Range)

   $R=u\times t$

   
   

   $R=15\times 0.5477=8.2155\text{m}$

   .
   Approx 8.22m.

    

Question 39

Question: Two sound waves have frequencies of 12Hz and 10Hz. Calculate their beats period.

• (a) 2.0s

• (b) 1.2s

• (c) 1.0s

• (d) 0.5s

Topic: Sound (Beats)
Correct Option: (d)
Detailed Explanation:
Beat Frequency (

fbf_bfb​

) =

f1−f2=12−10=2Hzf_1 - f_2 = 12 - 10 = 2\text{Hz}f1​−f2​=12−10=2Hz

.
Beat Period (

$T$

) =

1fb=12=0.5s\frac{1}{f_b} = \frac{1}{2} = 0.5\text{s}fb​1​=21​=0.5s

.

Question 40

Question: A lens that is thinner at the middle and thicker at the edges is (a)

• (a) diverging

• (b) converging

• (c) plano-convex

• (d) converging meniscus

Topic: Lenses
Correct Option: (a)
Detailed Explanation:
This is the physical definition of a Concave lens. Concave lenses diverge light rays, so they are also called Diverging lenses. (Convex/Converging lenses are thicker at the middle).