CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Two particle held at different height $$a$$ and $$b$$ above the ground are allowed to fall from rest. The ratio of their velocities on reaching the ground is


A
a:b
loader
B
a:b
loader
C
a2:b2
loader
D
b:a
loader

Solution

The correct option is B $$\sqrt{a}:\sqrt{b}$$
Let we assume that both particles takes time $$t_1$$ & respectively to reach the ground.

$$S = ut + \dfrac{1}{2} at^2$$
$$g = 10 \,m/s$$
For first particle, $$a = 5t_1^2$$   ...(1)
For second particle, $$b = 5t_2^2$$   ...(2)

By dividing $$eq^n (1)/ eq^n (2)$$

$$\dfrac{t_1}{t_2} = \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$$

Now applying, $$V = u + at$$
$$10 t_1 = V-1$$    ...(3)
$$10 t_2 = V_2$$   ...(4)

Dividing $$eq^n (3)/(4)$$
$$\dfrac{V_1}{V_2} = \dfrac{t_1}{t_2} = \dfrac{\sqrt{a}}{\sqrt{b}}$$

$$V_1 : V_b = \sqrt{a}:\sqrt{b}$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image