KVPY-SA 2018 Physics Paper with Solutions

Question 1: A block of wood is floating on water at 0°C with volume V₀ above water. When the temperature of water increases from 0 to 10°C, the change in the volume of the block that is above water is best described schematically by the graph

Answer: a

Change in volume can occur due to following reason:

(i) Density/volume of fluid changes due to increase or decrease in temperature.

(ii) The volume of material which is immersed changes due to temperature change [This argument need not to be considered as change in volume of the material is infinitesimal as co-efficient of volume thermal expansion has a very low value]

• For floating condition we can write [T = O°C] [Archimedes – principle]

Weight of the block = Buoyant force on the block

W = F_B

W = ρ_watergV_immersed ……(1)

• As buoyant force depend on the density of water if it changes due to temperature change [ρ_water is maximum at 4°C]

From (1)

mg =ρ_waterg V_immersed [m = mass of block]

V_immersed= m/ρ_water

Volume outside water:-

V_o = V_total – V_immersed

0 – 4°C Density ↑ V_immersed ↓ V_o ↑

4°C – 10°C Density ↓ V_immersed ↑ V_o↓

Hence volume outside Block changes according to immersed volume which depend on density of water.

Above statement written to understand that density variation curve & volume outside water of the block with temperature change is same.

Question 2: A very large block of ice of the size of a volleyball court and of uniform thickness of 8m is floating on water. A person standing near its edge wishes to fetch a bucketful of water using a rope. The smallest length of rope required for this is about

a. 3.6 m
b. 1.8 m
c. 0.9 m
d. 0.4 m

Answer: c

From Archimedes principle:-

Weight of ice-Block = Buoyant force on block

W = F_B

ρ_i. gV = ρ_w g _Vimmersed

We know that

Density of ice, ρ_i = 0.9 gm/cm³

Density of water

ρ_w = 1 gm/cm³

H_i = 0.9H

H_i = 7.2 m

So Height outside water:

H_o=8 – 7.2

H_o = 0.8m

Nearest value is 0.9m

Hence, the smallest length of rope required is 0.9m.

Question 3: A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show

a. the water coming down forming a parabolic stream
b. the water going up forming a parabolic stream
c. the water coming out in a straight line
d. no water coming out

Answer: d

When a filled box with water with small hole falls freely, then pressure is same at inside and outside of hole i.e. atmospheric pressure. No water will come out of hole as there is no pressure difference.

Option (d) is correct

Question 4: An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries 4 kg of water and the rate of evaporation is 20 g per hour, temperature of water in it decreases by ΔT in two hours. The value of ΔT is close to (ratio of latent of evaporation to specific heat of water is 540 °C)

a. 2.7 °C
b. 4.2 °C
c. 5.4 °C
d. 10.8°C

Answer: c

As evaporation occur the temperature of the water decreases as it is a cooling process

Rate of evaporation of water = 20 g per hour

Question 5: Two plane mirrors are kept on a horizontal table making an angle θ with each other as shown schematically in the figure. The angle θ is such that any ray of light reflected after striking both the mirrors return parallel to its incident path. For this to happen, the value of θ should be

a. 30°
b. 45°
c. 60°
d. 90°

Answer: d

Angle of deviation [δ]:-

δ = 360°– 2θ

given:- δ = 180°

180° = 360° – 2θ

θ = 90°

Hence; option (d) is correct

Question 6: A certain liquid has a melting point of –50°C and a boiling point of 150°C. A thermometer is designed with this liquid and it’s melting and boiling points are designated as 0° L and 100°L. The melting and boiling points of water on this scale are

a. 25° L and 75° L, respectively
b. 0° L and 100° L, respectively
c. 20° L and 70° L, respectively
d. 30° L and 80° L, respectively

Answer: a

This question is based on calibration of thermometer on various temperature scale in which the value of two standard reference point are given i.e. melting and boiling points.

Now for water:-

Alternate solution:-

On Comparison

1°L = 2°C

Question 7: One can define an alpha–Volt (α V) to be the energy acquired by an α particle when it is accelerated by a potential of 1 Volt. For this problem you may take a proton to 2000 times heavier than an electron. Then

a. 1 ∝ V = 1 eV/4000
b. 1 ∝ V = 2eV
c. 1 ∝ V = 8000 eV
d. 1 ∝ V = 1 eV

Answer: b

From work kinetic – energy theorem:

Work done= Change in kinetic energy

W = ΔK

Work done =q [ΔV]

q = charge

ΔV = potential difference

q_α = 2qe

For same –potential:

ΔK_α = 2ΔKe

1αV = 2eV

Hence, option (B) is correct

Question 8: In a particle accelerator, a current of 500 μA is carried by a proton beam in which each proton has a speed of 3 × 10⁷ m/s. The cross sectional area of the beam is 1.50 mm². The charge density in this beam in coulomb/m³ is close to–

a. 10^–8

b. 10^–7

c. 10^–6

d. 10^–5

Answer: d

From current Electricity:-

i = neAVd

n = no. of electrons per unit volume

A = cross-sectional area

V_d = Drift speed

Charge-density = q/V

q/V= ne = charge on particle per unit volume [charge density]

ne = i/AV_d

= 10^–5 c/m³

Hence option (D) is correct.

Question 9: Which of the following is NOT true about the total lunar eclipse?

a. A lunar eclipse can occur on a new moon and full moon day.
b. The lunar eclipse would occur roughly every month if the orbits of earth and moon were perfectly coplanar
c. The moon appears red during the eclipse because the blue light is absorbed in earth’s atmosphere and red is transmitted
d. A lunar eclipse can occur only on a full moon day

Answer: a

A lunar eclipse can occur only on full moon day.

Hence option (a) is not correct and this is the required answer.

Question 10: Many exoplanets have been discovered by the transit method, wherein one monitors a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius R and the parent star has radius 100R. If I₀ is the intensity observed on earth due to the parent star, then as the exoplanet transits

a. the minimum observed intensity of the parent star is 0.9 I₀

b. the minimum observed intensity of the parent star is 0.99 I₀

c. the minimum observed intensity of the parent star is 0.999 I₀

d. the minimum observed intensity of the parent star is 0.9999 I₀

Answer: d

Intensity ∝ [Area] ∝ [Radius]²

A_p = effective area of exoplanet

A_s = effective Area of star

I = 0.9999I₀

Question 11: A steady current I is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region

a. the current density decreases in value
b. the magnitude of the electric field increases
c. the current density remains constant
d. the average speed of the moving charges remains constant

Answer: b

Given :-

I = constant

Current Density

Hence option (A) & (C) are wrong due to above expression

Now

magnitude of Electric-Field increases. [Option B is correct]

I = neAV_d

Hence Drift speed/average speed increases along the length.

Option (D) is incorrect

Question 12: Select the correct statement about rainbow

a. We can see a rainbow in the western sky in the late afternoon
b. The double rainbow has red on the inside and violet on the outside
c. A rainbow has an arc shape since the earth is round
d. A rainbow on the moon is violet on the inside and red on the outside

Answer: b

Question 13: Remote sensing satellites move in an orbit that is at an average height of about 500 km from the surface of the earth. The camera onboard one such satellite has a screen on area A on which the images captured by it are formed. If the focal length of the camera lens is 50 cm, then the terrestrial area that can be observed from the satellite is close to

a. 2 × 10³A
b. 10⁶A
c. 10¹²A
d. 4 × 10¹²A

Answer: c

Area of Image = m² × area of object

A_i = m²A_o

m = magnification = Height of image/Height of object

=

V= 50 cm = 50 x 10^-2 m

u = 500 Km = 500 × 10³m

= 10¹² A

Question 14: Letters A, B, C and D are written on a cardboard as shown in the picture.

The cardboard is kept at a suitable distance behind a transparent empty glass of cylindrical shape. If the glass is now filled with water, one sees an inverted image of the pattern on the cardboard when looking through the glass. Ignoring magnification effects, the image would appear as

Answer: d

Given:-

Lens is of cylindrical shape and we know the image formed by cylindrical lens is Erect & Laterally inverted.

Hence option (d) is correct Answer

Question 15: If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity profile as function of height? Assume g = 10 m/s².

Answer: a

From Newton’s Third equation of motion

V² = u² + 2as

Here:- a = – g

At ,S = H_max = h

V = 0

h = 101.25 m = 101 m

As h = V²/2g

at s = 0 ; V = 45m/s

at s = h ; V = 0

Hence; option (A) is correct

Question 16: A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t₀ seconds to make coffee at 70°C when the initial temperature of the mixture is 25°C. The value of t₀ is close to (ratio of latent heat of evaporation to specific heat of water is 540°C and specific heat of the mixture can be taken to be the same as that of water)

a. 30
b. 45
c. 60
d. 90

From principle of calorimetry:

Heat provided by steam = Heat gain by mixture

For Steam

Q_steam = Q_mixture

M_sL + MsS Δ θs = M_m SΔ θm

From the definition of Latent Heat = M_sL

From the definition of Specific Heat = M_sSΔ θs

From the definition of Specific Heat = M_m SΔ θm

Substituting the values

Question 17: A person in front of a mountain is beating a drum at the rate of 40 per minute and hears no distinct echo. If the person moves 90 m closer to the mountain, he has to beat the drum at 60 per minute to not hear any distinct echo. The speed of sound is

a. 320 ms^–1

b. 340 ms^–1

c. 360 ms^–1

d. 380 ms^–1

Answer: c

For not hearing Echo, time interval between the beats of drum must be equal to the time of Echo.

t₁= 2x/V

Now,

f₂ = 60 per minute

f₂ = 2[x-90]/V

V= 360 m/s

Question 18: A glass beaker is filled with water up to 5 cm. It is kept on top of a 2 cm thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, it’s apparent depth from the water surface is d cm. Value of d is close to (the refractive indices of water and glass are 1.33 and 1.50, respectively)

a. 2.5
b. 5.1
c. 3.7
d. 6.0

Answer. b

Substituting the values of µ_w and µ_g

=

=

d=5.1cm

Question 19: A proton of mass m and charge e is projected from a very larger distance towards an α particle with velocity v. Initially, α particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be

a. e²/4 π ε₀mv²

b. 5e²/4 π ε₀mv²

c. 2e²/4 π ε₀mv²

d. 4e²/4 π ε₀mv²

Answer: d

From conservation of linear momentum [As ∑F = 0]

P_initial = P_final

mv = [m + 4 m] V’

v = 5V’

V’ = v/5

Now from conservation of mechanical energy:-

Method II

Applying conservation of mechanical Energy with respect to centre of mass frame:-

= potential energy at minimum separation d_min

Where, 2e=charge of α particle, e=charge of proton

And K=Coulomb’s constant=1/4πε₀

μ = Reduced Mass = m₁m₂/m₁+m₂

Where, m₁= mass of α-particle and m₂= mass of proton

Question 20: A potential is given by V(x) = k(x + a)²/2 for x < 0 and V(x) = k(x – a)²/2 for x > 0. The schematic variation of oscillation period (T) for a particle performing periodic motion in this potential as a function of its energy E is

Answer: b

For small energy; particle oscillates in one well either it will oscillate about x = +a OR x = –a & Time period of this oscillation is T. [Below energy Level (1/2)Ka² ]

When this particle gain energy equal to the [(1/2)Ka² ] it will oscillate in both well so total time period now become 2T so there is a sudden jump in time period.

Now as you increase energy more than [(1/2)Ka² ] then:

Hence time period starts decreasing from 2T. Hence option B is correct.

Now for high energy we can conclude that curve gets symmetrical and time period again asymptotically approaches T.