# [Solved] JEXPO 2018 Physics Question Paper

### [Solved] JEXPO 2018 Physics Question Paper

JEXPO 2018 Chemistry Question Paper with Solution

**Q.1)** What is the angle between incident and reflected rays when a ray of light is incident normally on a plane mirror?

90°

45°

0°

180°

## Answer

Correct Answer: 180°

## Explanation

When ray of light incident normally on a plane mirror, then incident angle = 0°.

∴ Reflection angle also be 0° [according to the 2nd law of reflection of light].

Again, we know that in that case direction of incident ray and reflected ray are opposite to each other.

Hence, the required angle must be 180°.

**Q.2)** Which mirror has a wider field of view?

cylindrical mirror

concave mirror

plane mirror

convex mirror

## Answer

Correct Answer: convex mirror

## Explanation

Convex mirror because a mirror is diverging in nature. Light rays incident at its extreme corners gets diverged following laws of reflection. Hence if we draw the diagram we can clearly understand that the reflected rays follow a wider path. While in case of a plane mirror the divergence is not so large.

**Q.3)** An object of size **2.0 cm** is placed perpendicular to the principal axis of a concave mirror. The distance of the object from the mirror equals the radius of curvature. The size of the image will be

1.5 cm

0.5 cm

1.0 cm

2.0 cm

## Answer

Correct Answer: 2.0 cm

## Explanation

Size of the image must be same as the object size if an object is placed perpendicular to the principal axis of a concave mirror at a distance equal to **2 f** or

*(radius of curvature).*

**r**

**Q.4)** On a rainy day, small oily films on water show brilliant colours. This is due to

scattering

interference

dispersion

polarisation

## Answer

Correct Answer: interference

## Explanation

No explanation

**Q.5)** Variation of resistance of semiconductor with temperature is represented by which graph?

## Answer

## Explanation

For semiconductor, resistance decreases with increases the temperature.

**Q.6)** How much work is done in moving **4C** across two points having potential difference **10V**?

40J

2.5J

0.4J

10J

## Answer

Correct Answer: 40J

## Explanation

Work-done in moving a electrical charge, W = Potential difference (V) × amount of charge (Q) ⇒ 10 × 4 Joules = 40 J

**Q.7)** Rectifier is used to convert

AC to DC

low voltage to high voltage

DC to AC

None of those

## Answer

Correct Answer: AC to DC

## Explanation

No explanation.

JEXPO 2018 Chemistry Question Paper with Solution

**Q.8)** If a **24V** and a **10V** battery series opposing, total voltage is

34V

10V

24V

14V

## Answer

Correct Answer: 14V

## Explanation

According to the question, both the batteries are connected in opposite direction.

Thus, total voltage = (24 − 10) V ⇒ 14 V

**Q.9)** What is the equivalent resistance between **A** and **B**?

3 Ω

2 Ω

4 Ω

1 Ω

## Answer

Correct Answer: 2 Ω

## Explanation

4Ω and 2Ω are connected in series [see figure].

∴ Equivalent resistance of 4Ω and 2Ω resistances = (4 + 2)Ω ⇒ 6Ω.

Again 6Ω and 3Ω resistances are connected in parallel combination.

Thus, the required equivalent resistance between A and B =

6×36+3=2 Ω

**Q.10)** A car runs on a circular track at a constant speed. The circular track has a radius of 100 *m*. If the car takes 62.8 *seconds* on each lap, it’s average speed is

20 ms^{−1}

5 ms^{−1}

10 ms^{−1}

Zero

## Answer

Correct Answer: 10 ms^{−1}

## Explanation

By the definition of average speed, we have

Average speed =

total distance total time⇒

circumference of circular track total time⇒

2×3.14×100 62.8ms^{−1} = 10 ms^{−1}

**Q.11)** Two persons start running towards each other from two points that are 120 *m* apart. First person runs with a speed of 5 *m/s* and the other with a speed of 7 *m/s*. Both the persons meet after

10s

24s

48s

1 min

## Answer

Correct Answer: 10s

## Explanation

Let, 1st person covered S_{1} m distance in time “t” seconds while 2nd person covered (120 − S_{1}) m in “t” seconds.

By the mathematical relation of speed, distance and time, we have

Distance = Speed × Time

∴ S_{1} = 5t

For 2nd person, 120 − S_{1} = 7t

Now, substituting the value of S_{1}, we have

⇒ 120 − 5t = 7t

⇒ 12t = 120

∴ t = 10 seconds.

**Q.12)** A stone is dropped into a well in which the level of water is 8 *m* below the top of the well. If ** v** is the velocity of sound in air, the time after which the splash is heard will be given by

2H g + H v

2H v

H 2g + 2H v

2Hv+Hg

## Answer

Correct Answer:

2H g + H v

## Explanation

The required time = time to fall a stone from the top of well to surface of water (T_{1}) + time to cover H m by sound (T_{2})

Calculation of T_{1}

By the formulas of kinematics, we have

(V_{stone})^{2} = (u_{stone})^{2} + 2gH

⇒ 2gH [∵ initial velocity of stone, u_{stone} = 0]
∴

V stone = 2gHNow, T_{1} =

V stone g⇒

2gH g=

2H g[ using V = u + gt]
Calculation of T_{2}

We have, Distance = Speed × Time.

∴ T_{2} =

H vHence, the required total time = T_{1} + T_{1} ⇒

2Hg+Hv

**Q.13)** If time-displacement graph of a particle is parallel to its time axis, velocity of the particle is

infinity

equal to acceleration

unity

zero

## Answer

Correct Answer: zero

## Explanation

By the definition of velocity (i.e. instantaneous velocity), we have

Instantaneous Velocity =

Change in Displacement Change in Time

According to the question, displacement is constant with respect to time [see graph].

∴ Change in displacement w.r.t. time = 0.

Hence, Velocity =

0 Change in Time

⇒ 0 (zero).

**Q.14)** **M***gm* of ice at 0°C is to be converted to water at 0°C. If * L* is the latent heat of fusion of ice, the quantity of heat required for the above operation would be

*ML* calorie

M L

calorie

None of those

LM

## Answer

Correct Answer: *ML* calorie

## Explanation

Heat required to change the state of matter, Q = mass of substance (M) × latent heat (L)

∴ Q = ML calories. (Ans.)

**Q.15)** When it is raining, dew point is

room temperature

50°C

100°C

0°C

## Answer

Correct Answer: 0°C

## Explanation

No explanation.

**Q.16)** How much ice must be added to 100 *gm* of water at 30°C in order to reduce its temperature to 20°C?

10 g

400 g

80 g

None of those

## Answer

Correct Answer: 10 g

## Explanation

Let, ‘m’ gm ice at 0°C must be added with 100 gm of water at 30°C.

According to the principle of Calorimetry, we have

Total heat released = Total heat gained.

∴ Heat released by water at 30°C to 20°C = Heat gained by ice at 0°C to water at 0°C + Heat gained by water at 0°C to 20°C

100 × 1 × (30 − 20) = m × 80 + m × 1 × (20 − 0) [Latent heat of fusion of ice = 80 cal/gm and sp. heat of ice = sp. heat of water = 1 ]
1000 = 80m + 20m

∴ m = 10 gm.

**Q.17)** Which of the following substances has greatest specific heat?

Iron

Water

Mercury

Copper

**Q.18)** A bottle of water at 0°C is opened on the surface of the Moon. What will happen?

No change will occur

Water freezes

Water decomposes into O_{2} and H_{2}

Water will boil

## Answer

Correct Answer: Water freezes

## Explanation

We know that boiling point of substances get decreases with decrease of pressure.

Since, the atmospheric pressure is very low in the moon. Thus, when the lid of water bottle is opened the boiling point of water gets fall very fast and suddenly water will boil and due to certain change in state of water from liquid to vapor, it will take latent heat from the vapor or bottle or both and hence it will convert into ice. [I hope it is clear to all]

**Q.19)** Freezing mixture is a mixture

which freezes at 0°C

which solidifies water

which produces very low temperature

which is used in medicine

## Answer

Correct Answer: which produces very low temperature

## Explanation

No explanation.

**Q.20)** The number of joules contained in **1 kWh** is

36 × 10^{2}

36 × 10^{3}

3.6 × 10^{6}

36 × 10^{4}

## Answer

Correct Answer: 3.6 × 10^{6}

## Explanation

1 kWh = 1000 watts × 1 hour

⇒ 1000 watts × 3600 seconds

⇒ 3.6 × 10^{6} Joules. [ 1 Joule = 1 watt × 1 second]

**Q.21)** In case of negative work, the angle between force and displacement is

180°

270°

90°

0°

## Answer

Correct Answer: 180°

## Explanation

When direction of displacement(s) is opposite to the direction of applied force (F), then work-done is said to be negative work-done.

Now, by the mathematical equation of work-done (W), we have

W = F × S × cosθ [ θ = angle b/w force and displacement]
Now, for θ = 180° , cosθ = − 1

Hence, W ⇒ −ve.

**Q.22)** 3730 ** watts** = _____

**H.P.**2

5

746

6

## Answer

Correct Answer: 5

## Explanation

We have, 1 H.P. (Horse Power) = 746 watts.

Hence, 3730 watts =

3730746=5 H.P.

**Q.23)** An automobile engine of mass * m* accelerates and a constant power

*is applied by the engine. The instantaneous speed of the engine will be*

**P**Pt 2m

( Pt 2m ) 1 2

2Pt m

(2Ptm)12

## Answer

Correct Answer:

( 2Pt m ) 1 2

## Explanation

We have, instantaneous power, P = Force × Velocity

⇒ P = (ma)v [ Force = mass (m) × acceleration (a); and v = velocity]
P =

m( dv dt )vIntegrating both sides, we have

⇒

∫ P m dt= ∫ vdv⇒

P m ( t )= v 2 2∴ v =

(2Ptm)12

**Q.24)** If K.E. of a body increases by **0.1%**, the present increase in momentum would be

0.05%

1%

0.1%

10%

## Answer

Correct Answer: 0.05%

## Explanation

Given that, % change in K.E., △E_{k}% = 0.1%

⇒

E k2 − E k1 E k1 ×100%=0.1%⇒

E k2 − E k1 E k1 = 0.1 100⇒

E k2 E k1 −1= 1 1000⇒

E k2 E k1 = 1001 1000⇒

( P 2 ) 2 2m ( P 1 ) 2 2m = 1001 1000 [∵ K.E., E_{k} = * ^{p2}⁄_{2m}* , where P = momentum and m = mass]
⇒

( P 2 ) 2 ( P 1 ) 2 = 1001 1000⇒

P 2 P 1 = 10010 10000⇒

P 2 P 1 = 100.05 100⇒

P 2 − P 1 P 1 ×100%=

100.05−100 100 ×100%⇒ △P % = 0.05%

∴ % Change in momentum (△P) = 0.05%

**Q.25)** A bomb of 12 *kg* explodes into two pieces of masses 4 *kg* and 8 *kg*. The velocity of 8 *kg* is 6 *m/s*. The kinetic energy of the other mass is

32J

24J

128J

288J

## Answer

Correct Answer: 288J

## Explanation

By the **Conservation law of momentum**, we have

m_{1}v_{1} = m_{2}v_{2} [ where, m_{1} = 4 kg, m_{2} = 8 kg, v_{1} = velocity of 4 kg mass and v_{2} = 6 m/s ]
⇒ v_{1} =

m 2 v 2 m 1⇒

8×6 4 =12 m/sHence, K.E. of 4 kg mass =

1 2 ×4× ( 12 ) 2⇒ 288 Joules. (Ans.)

**Q.26)** A raised hammer possesses

K.E. only

Electrical energy

Gravitational P.E.

Sound energy

## Answer

Correct Answer: Gravitational P.E.

## Explanation

No explanation.

**Q.27)** How many fundamental units are present in the SI system of units?

5

6

8

7

**Q.28)** The ratio of one micron to one nanometer is

10^{−6}

10^{−3}

10^{6}

10^{3}

## Answer

Correct Answer: 10^{3}

## Explanation

1 micron = 10^{−6} meter and 1 nanometer = 10^{−9} meter

∴ The required ratio =

10 −6 10 −9⇒ 10^{−6 + 9} = 10^{3}.

**Q.29)** One **second** is defined as _____ part of a mean **solar day**.

1 86,400

1 20,000

1 96,400

130,000

## Answer

Correct Answer:

1 86,400

## Explanation

By the definition of **mean solar day**, we have

1 mean solar day = (24 × 60 × 60) seconds ⇒ 86,400 seconds.

Hence, 1 second =

1 86,400part of ** mean solar day**.

**Q.30)** Relative density of mercury is

1

0.8

2.5

13.6

## Answer

Correct Answer: 13.6

## Explanation

No explanation.

**Q.31)** The weight of a body at the centre of earth is

infinite

equal to its mass

maximum

zero

## Answer

Correct Answer: zero

## Explanation

We have, weight = (mass) × (acceleration due to gravity)

Since, Acceleration due to gravity at the centre of earth is zero (0).

Hence, the required weight of any body at the centre of earth = 0 (zero).

**Q.32)** A stone dropped from the roof of a building takes 4 *second* to reach the ground. The height of the building is

9.8 m

39.2 m

19.6 m

78.4 m

## Answer

Correct Answer: 78.4 m

## Explanation

Using the kinematics formula, we have

H=ut+ 1 2 g t 2⇒

H= 1 2 g t 2 [∵ initial velocity of stone (u) = 0]
⇒ H = 0.5 × 9.8 × (4)^{2} [∵ g = 9.8 m/s^{2}]
∴ H = 78.4 m.

**Q.33)** A ball is thrown up and attains a maximum height of 100 *m*. It’s initial speed was

9.8 ms^{−1}

44.2 ms^{−1}

none of those

19.6 ms^{−1}

## Answer

Correct Answer: 44.2 ms^{−1}

## Explanation

The maximum height that can attain by a ball, u =

2gH [ where, initial velocity = u ; g = 9.8 m/s^{2} and H = 100 m]
⇒ u =

2× 9.8× 100⇒ 44.27 m/s.

**Q.34)** Two planets have the same density but different radii. The acceleration due to gravity would be

nothing can be decided.

greater on the larger planet.

same on both planets.

greater on the smaller planets.

## Answer

Correct Answer: greater on the larger planet.

## Explanation

Since, g =

4 3 πGRρ [ where, G = gravitational constant, R = radius, ρ = density ]

g=( 4 3 πGρ )R⇒ g = (constant) × R [ since, G, ρ and π are constant] Hence,

g α R∴ Acceleration due to gravity (g) would be greater on greater planet (i.e. greater radius).

**Q.35)** If the earth suddenly strings by **one-third** of its present radius, acceleration due to gravity would be

4 9 g

3 2 g

9 4 g

g3

## Answer

Correct Answer:

g 3

## Explanation

We have,

g 1 g 2 = r 1 r 2⇒

g 2 = g 1 × r 2 r 1⇒

g 2 = g 1 × ( r 1 3 ) r 1 [ ∵

r 2 = r 1 3] ⇒

g 2 = g 1 3Now, substituting g_{1} as g, we have

∴ The required acceleration due to gravity =

g3

**Q.36)** Elements show radioactivity when the ratio of neutrons to protons in the nucleus of an atom exceeds

1.5

1.6

2.5

1.7

## Answer

Correct Answer: 1.5

## Explanation

No explanation.

**Q.37)** The highest binding energy per nucleon is found in

C 28 60 o

P 82 210 b

U 92 235

F2656e

## Answer

Correct Answer:

F 26 56 e

## Explanation

No explanation.

**Q.38)** Temperature at the centre of Sun is equivalent to

20 million Kelvin

50 million Kelvin

30 million Kelvin

40 million Kelvin

## Answer

Correct Answer: 20 million Kelvin

## Explanation

No explanation.

**Q.39)** Radiations bend in opposite direction in magnetic field are

α and β – radiations

γ and β – radiations

α and γ – radiations

all radiations

## Answer

Correct Answer: α and β – radiations

## Explanation

We know that α and β be charged particles. Hence, α and β radiations get deflected by the presence of magnetic field and it also be affected by the electric field.

**Q.40)** The volume of a gas sample is increased while its temperature is held constant. The gas exerts a lower pressure on the walls of container partly because its molecules strike the walls

with less force

with lower velocities

with less energy

less often

## Answer

Correct Answer: less often

## Explanation

The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert small force on the wall. The pressure exerted by the gas is due to the sum of all these collision forces. The more particles that hit the walls, the higher the pressure.

If a gas is heated up, its particles move around more quickly. They hit the walls of their container harder and more often. This increases the pressure.

**Q.41)** How many moles of **H** atoms are present in 1 *mole* of water, **H _{2}O**?

1

^{2}⁄_{3}

2

3

## Answer

Correct Answer: 2

## Explanation

1 mole H_{2}O contains 1 mole oxygen atoms and 2 moles hydrogen atoms.

**Q.42)** An oxygen molecule has 16 times the mass of a hydrogen molecule. A sample of hydrogen gas whose molecules have same average K.E. as the molecules in a sample of oxygen at 400 K is at a temperature of

25 K

1600 K

400 K

6400 K

## Answer

Correct Answer: 6400 K

## Explanation

Kinetic energy (K.E.) of 1 gm gas =

3 k B T 2M where,

k_{B} = Boltzmann’s constant

R = gas constant

M = mass of each molecule.

Now, according to the problem, we have

( E o ) gram = ( E H ) gram⇒

3 k B T o 2 M o = 3 k B T H 2 M H⇒

T o = T H × M o M H⇒

T o = 400×16 M H M H [ since,

M o =16 M H] Hence,

To=6400 K

**Q.43)** Wind energy is generated when power of wind must be

> 5 km/h

> 10 km/h

> 15 km/h

= 10 km/h

## Answer

Correct Answer: > 10 km/h

## Explanation

No explanation.

**Q.44)** The radiation in sunlight that gives us feeling of hotness is

infra-red

visible radiation

red

ultra-violet

## Answer

Correct Answer: infra-red

## Explanation

No explanation.

**Q.45)** Which of the following causes the least pollution when burnt?

Petrol

Coal

Diesel

Natural gas

## Answer

Correct Answer: Natural gas

## Explanation

Among all the given fuels, **natural gas** emits the **least** amount of **carbon-di-oxide** into the air during the combustion.

**Q.46)** To hear a distinct eco each time interval between the original sound and the reflected sound must be

1s

0.2s

2s

0.1s

## Answer

Correct Answer: 0.1s

## Explanation

To hear a distinct eco each time interval between the original sound and the reflected sound must be

110 s⇒0.1 s

**Q.47)** Loud sound can travel a larger distance, due to

higher amplitude

higher frequency

higher energy

higher speed

## Answer

Correct Answer: higher amplitude

## Explanation

No explanation.

**Q.48)** Infra-sound can be heard by

Rhinoceros

Bats

Man

Dolphins

## Answer

Correct Answer: Rhinoceros

## Explanation

No explanation.

**Q.49)** The time period of a vibrating body is 0.05*s*. The frequency of waves it emits is

2 Hz

20 Hz

200 Hz

5 Hz

## Answer

Correct Answer: 20 Hz

## Explanation

By the relation of frequency and time period, we have

f= 1 T where,

f= frequency of wave

T = time period.

⇒

f= 1 0.05 Hz⇒ 20 Hz

**Q.50)** The image formed by a plane mirror is always

real and erect

real and inverted

virtual and erect

virtual and inverted

## Answer

Correct Answer: virtual and erect

## Explanation

No explanation.

JEXPO 2018 Chemistry Question Paper with Solution

## Thank You

☛ *If you still have any query, then feel free to contact ☎ or comment ✍ me.*

* Thank you.*

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