# Important MCQs for NEET examination

In this particular article Important MCQs for NEET examination, we will be going to discuss important questions or neet examination along with their answer.

## Part A

### 1)

A ball is thrown vertically upwards in the air and reaches a maximum height of 50m. The velocity of projection of ball is (g=9.8m s)

(a) 45 m s⁻¹

(b) 31 m s⁻¹

(c) 10 m s⁻¹

(d) None of these

Answer: (b)

Given h= 50m

a = -g = -9.8 and ν=0.

For vertically upward motion

ν² = u² – 2gh

u = √2gh = √19.6×50 ≅ 31.30 m s⁻¹

### 2)

A train is moving along a straight path with uniform acceleration. Its engine passes across a pole with a velocity of 60 km h⁻¹ and the end (guaed’s van) passes across the same pole with a velocity of 80 km h⁻¹. The middle point of the train will pass across the same pole with a velocity

(a) 70 km h⁻¹

(b) 70.7 km h⁻¹

(c) 65 km h⁻¹

(d) 75 km h⁻¹

Answer: (b)

From ν² – u² = 2as,

(80)²-(60)²/2a = s

s = 6400-3600/2a = 1400/a

The middle point of the train is to cover a distance

s/2 = 700/a

From ν² – u² = 2as

ν²- (60)² = 2a × 700/a = 1400

ν²= 1400 + 3600

ν= √5000 = 70.7 km h⁻¹

### 3)

What is the angle between P and the resultant of (P+Q) and (P-Q)?

(a) Zero

(b) tan¯¹(P/Q)

(c) tan¯¹(Q/P)

(d) tan¯¹(P-Q)/(P+Q)

Answer: The resultant of (P+Q) and (P-Q)

= P+Q + P- Q = 2P

Since 2P is parallel to P, hence the angle between them is zero.

### 4)

A projectile is thrown a²t an angle of θ= 45° to the horizontal, reaches a maximum height of 16m. Then choose the incorrect option.

(a) Its velocity at the highest point is zero.

(b) Its range is 64 m.

(c) Its range will decreases when it is thrown at an angle of θ = 30º.

(d) Both (b) and (c) are correct.

Answer: (a)

The maximum height reached by the projectile,

H = u² sin2θ/g or 16 = u²×sin²45 °/19.6

∴ u= √16×2×19.6

Horizontal range of a projectile,

R= u² sin2θ/g = 16×2×19.6×sin90°/9.8 = 64m

When θ=45°, sin2θ=1

But when θ= 30°, sin2θ = √3/2= o.866 ; the range will decrease.

#### Part B

##### 5)

A disc of mass 100 g is kept floating horizontally in the air by firing bullets, each of mass 5 g with the same velocity at the same rate of 10 bullets per second. The bullets rebound with the same speed in opposite direction, the velocity of each bullet at the time of impact is

(a) 196 cm s⁻¹

(b) 9.8 cm s⁻¹

(c) 98 cm s⁻¹

(d) 980 cm s⁻¹

Answer: (d)

At the time of impact, 2m’vn = mg

∴ ν= mg/2m’n = 100×980/2×5×10 = 980 cm s⁻¹

##### 6)

A ball of mass is thrown vertically upwards with a velocity v. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower is

(a) v√mg/mg+F

(b) v√F/mg + F

(c) v√mg-F/mg+F

(d) v√mg+F/mg

Answer: (c)

For an upward motion,

retarding force= mg + F

∴ Retardation, a = (mg + F)/m

and distance, s= v²/2a = v²m/2(mg+F)

For the downward motion,

net force = mg- F

∴ Acceleration, a’=(mg-F)/m

As s=s’ ∴v’=v√(mg-F)/(mg+F)

##### 7)

A Cyclic ride up a hill with a constant velocity. If the length of the connection rod of the pedal is r=25 cm, the time of revolution of the rod is t= 2 s and the mean force exerted by his foot on the pedal is F=15 kg wt. The power developed by the cyclist is

(a) 1154 W

(b) 115.4W

(c) 15W

(d) 11.5W

Answer: (b)

The velocity of the cyclist,

v = rω = r(2π/t) = (1/4)×2π/2 = (π/4) m s

The power developed by the cyclist,

P = F×v = (15×9.8)×π/4 = 115.4 W

##### 8)

The angle of contact between glass and water is 0° and it rises in a capillary up to 6 cm when its surface tension is 70 dyne cm⁻¹. Another liquid of surface tension 140 dyne cm⁻¹, the angle of contact 60° and relative density 2 will rise in the same capillary by

(a) 12 cm

(b) 24 cm

(c) 3 cm

(d) 6 cm

Answer: (c)

Capillary rise, h= 2Scosθ/rρg

Because of h₂/h₁=S₂/S₁ ×cosθ₂/cosθ₁ ×ρ₁/ρ₂×r₁/r₂

Or h₂/h₁ = 140/70 ×cos60⁰/cos0⁰× 1/2 ×1= 1/2

Or h₂=h₁/2 = 3cm

##### 9)

Two balls of masses m₁ and m₂ are separated from each other and a charge is placed between them. The whole system is at rest on the ground. Suddenly, the charge explodes and the masses are pushed apart. Mass m₁ travels a distance s₁ and then it stops. If the coefficient of friction between the balls ground is same, mass m₂ stops after covering a distance

(a) s₂=(m₁/m₂)s₁

(b) s₂= (m₂/m₁)s₁

(c) s₂= (m₁²/m₂)s₁

(d) s₂= (m₂²/m₁²)s₁

Answer: From the conservation of momentum, we get m₁v₁= m₂v₂…….(1)

Also, 1/2 (m₁v₁²) = f₁s₁ = μm₁gs₁ …..(2)

and 1/2 (m₂v₂²) = f₂s₂ = μm₂gs₂ …..(3)

Where μ= coefficient of friction

Dividing eqn (2) by eqn (3)

m₁v₁²/m₂v₂² = m₁s₁/m₂s₂ and v₁/v₂=m₂/m₁ [From equation 1]

Thus, v₁²/v₂² = s₁/s₂ or m₂²/m₁² = s₁/s₂ or s₂=m₁²s₁/m₂²

##### 10)

When the volume of an ideal gas is increased two times and the temperature is decreased half of its initial temperature, then pressure becomes

(a) 2 times

(b) 4 times

(c) 1/4 times

(d) 1/2 times

Answer: The ideal gas equation, PV= nRT

or P= nRT/V

where P is the initial pressure of the gas.

Given, V’=2V and T’=T/2

∴P’=(nRT/2)/2

Dividing equation (2) by equation (1)P’ =P/4