\(52K J\)
\(51K J\)
When force remains constant and motion occurs along a straight path, work is calculated as the product of force and distance traveled. However, when force varies within the distance traveled, calculus aids in resolving the problem by breaking it down into infinitesimal steps where force can be considered constant.
Consequently, the work accomplished by a variable force is expressed as,
\(W = ∫F dx\)
Here, the force is described by \(F = 5x\) N, and the displacement ranges from \(x = 2\) m to \(x = 4\) m.
Now the work performed,
\(W=∫Fdx\)
\(W=∫5x dx\)
\(W= 5∫_2^4 x dx\)
\(W = 5[\frac {x^2}{2}]_2^4\)
\(W=\frac 52[4^2-2^2]\)
\(W= 5\times 6\)
\(W=30\ J\)
So, the correct option is (B): \(\left(\frac{105}{2}K\right) J\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,



What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
Read More: Work and Energy