The work done in extending a spring by 40 mm is 1.52J. Calculate the elastic constant of the spring.
38.8 Nm-1
60.8 Nm-1
190.0 Nm-1
1900.0 Nm-1
3800.0 Nm-1
Correct answer is D
Force = Elastic constant x extension
Work done = \(\frac{1}{2} K \times e^2\)
Given Work done = 1.52J; Elastic constant (k) = ?; e = 40mm = 0.04 m
\(\therefore 1.52 = \frac{k \times (0.04)^2}{2}\)
\(3.04 = k \times 1.6 \times 10^{-3}\)
\(k = \frac{3.04}{1.6} \times 10^3\)
= 1.9 x 10\(^3\)
= 1900 Nm\(^{-1}\)
q2 ÷ 4πeod
q2 ÷ 4πeo
q ÷ 4πeod2
q ÷ 4πeod
q ÷ 4πeo2d
Correct answer is D
The electric potential at a point in an electric field is defined as the amount of work done in moving a unit positive charge from infinity to that point against the action of the field.
V = \(\frac{Q}{4 π∈r}\)
Both are inverted
The magnification of Y is greater than one while that of X is one
The magnification of Y is one while that of X is less than one
Y is inverted while X is erect
Y is real while X is virtual.
Correct answer is C
No explanation has been provided for this answer.
In the electrical method of making a magnet, the polar the magnet formed depends on the
amount of current passed
direction of the current
nature of the material used
orientation of the magnetic material in space
size of the magnetic material
Correct answer is B
No explanation has been provided for this answer.
zero
2.5
10
12.5
100
Correct answer is A
No explanation has been provided for this answer.