Physics Test 114

Result

Physics Test 114

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

The sum & differences of two perpendicular vectors of equal magnitudes are

A
also perpendicular and of equal lengths

B
also perpendicular and of different lengths

C
of equal length and have an obtuse angle between them

D
of equal lengths and have and acute angle between them

Solution
Question 2
4 / -1

An aeroplane is moving on a circular path with a speed of 250 kmph. The change in velocity (in kmph) in half revolution is

A
zero

B
250

C
500

D
250 √2

Solution
Question 3
4 / -1

If the sum of two unit vectors is also a unit vector, then the magnitude of their difference is

A
√2

B
√3

C
√5

D
√7

Solution
Question 4
4 / -1

A person aims a gun at a bird from a point, at a horizontal distance of 100 m.If gun can impart a speed of 500m/s to the bullet,at what height above the bird must he aim his gun in order to hit it? (g = 10 ms^-2)

A
10 cm

B
20 cm

C
50 cm

D
100 cm

Solution

From given the data

Range R = 2 ⨯100 m = 200 m

Initial velocity u = 500 m/s

We know that Range R =

Horizontal distance x = 100 m

Velocity v = 500 m.s^-1

Time taken to travel this distance t = x/v = 100/500 = 1/5s

Vertical distance travelled by the bullet in time 1/5s is given by,

If the person directly aims at the bird, the bullet will hit 20 cm below

the bird. Thus the gun must be aimed at 20 cm above the position of the bird.
Question 5
4 / -1

The hight y and distance x along the horizontal plane of a projectile on a certain planet (With no surrounding atmosphere) are given as y=y=8t-5t² mtr. and x = 6t metres, where t is in seconds. The velocity with which the projectile is projected is

A
8 m/s

B
6 m/s

C
10 m/s

D
unpredictable

Solution

Y = 8t - 5t²

differentiate wrt time

dy/dt = 8 -10t

we know, change in position per unit time is called , velcity .

e.g dy/dt = Vy

Vy = 8 -10t

now at t = 0 ( initially )

Vy =8 -10× 0 = 8

x = 6t

differentiate wrt time

dx/dt = 6

similarly , dx/dt = Vx =6

now,

V = Vx i + Vy j

= 8 i + 6 j

magnitude of velocity =| V | =√(8² + 6²) =10 m/s
Question 6
4 / -1

A helicopter flying at 20 m/s horizontally drops a rescue bag. Ignoring air resistance & taking g = 10ms^-2, the displacement of the bag 5 seconds after the drop began is nearly

A
100 m

B
125 m

C
160 m

D
225 m

Solution

S_x = 20 x 5

S^y = 1/2 x w x 25

S = √sx² + sy²

= 160.
Question 7
4 / -1

When the angular velocity of a uniformly rotating body has increased thrice, the resultant of forces applied to it increase by 60 N. Find the acceleration of the body in the two cases. The mass of the body, m = 3kg

A
2.5 ms^-2 , 7.5 ms^-2

B
7.5 ms^-2 , 22.5 ms^-2

C
5 ms^-2, 45 ms^-2

D
2.5 ms^-2 , 22.5 ms^-2

Solution
Question 8
4 / -1

In case of uniform circular motion which of the following physical quantities do not remain constant?

A
speed

B
kinetic energy

C
angular momentum

D
momentum

Solution

Though speed is constant but direction 0f motion and hence the velocity is variable. Therefore,the momentum is not constant
Question 9
4 / -1

A parachutist of weight W strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3g . force exerted on him by ground during landing is

A
1 W

B
2W

C
3W

D
4W

Solution

m = w/g

3mg = n - w

N = 4mg

= 4w
Question 10
4 / -1

An unloaded car moving with velocity u on a road can be stopped in a distance s. If passengers add 40% to its weight and braking force remains the same, the stopping distance at velocity u is now

A
1.4 s

B
√1.4 s

C
(1.4)²s

D
1/1.4 s

Solution