\(32ms^{-1} \)
\(50ms^{-1} \)
\(64ms^{-1} \)
\(100ms^{-1} \)
Correct answer is D
From Newton's second law:
\( F = \frac{mV - mµ}{t} \)
Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg
\( t = 0.8s \\
\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8} \\
\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1} \)
8.25 x 10-19J
5.28 x 10-19J
5.28 x 1019J
8.25 x 1019J
Correct answer is B
Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant
Energy emitted = hf
= 6.6 x 10-34 x 8.0 x 10-14
= 5.28 x 10-19J
0.03A
0.04A
0.05A
0.06A
Correct answer is C
From faraday's law of electrolysis
M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\
\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
30.0\(^{\circ} \)
37.2\(^{\circ} \)
42.0\(^{\circ} \)
48.6\(^{\circ} \)
Correct answer is B
For the equilateral glass prism A = 600
\( \frac{3}{2} \text{Sin} \frac{\left( \frac{Dm + 60}{2}\right)}{\left(\frac{60}{2}\right)}= \frac{\left( \text{ Sin }\frac{Dm + 60}{2} \right)}{0.5} \\
\implies Dm = 37.2^{\circ} \)
\( 2 \times 10^{-4}J \)
\( 2 \times 10^{-3}J \)
\( 2 \times 10^{1}J \)
\( 2 \times 10^{2}J \)
Correct answer is B
Work done by friction = Friction force x Displacement from the relation, f = uR
where u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)