If the momentum of a body increases by 0.01%, its kinetic energy will increase by

The correct option is 0.02 %.

Step 1: Establish the relationship between kinetic energy and momentum
The kinetic energy (K) of a body of mass m moving with velocity v is expressed as:

$$K=\frac{1}{2}m{v}^2$$
The momentum (p) of the body is defined as:

$$p=mv$$
Rearranging the momentum formula to solve for velocity gives:

$$v=\frac{p}{m}$$
Substituting this velocity term into the kinetic energy formula yields:

$$K=\frac{1}{2}m{\left(\frac{p}{m}\right)}^2=\frac{p^2}{2m}$$

Step 2: Relate small percentage changes using differentiation
Since the percentage change in momentum is extremely small (0.01%, which is far less than 10%), we can use the differential approximation method to find the corresponding change in kinetic energy.
Taking the natural logarithm ($ln$) on both sides of the kinetic energy relation:

$$ln\left(K\right)=ln\left(\frac{p^2}{2m}\right)$$
Applying logarithmic identities, this becomes:

$$ln\left(K\right)=2ln\left(p\right)-ln\left(2m\right)$$
Differentiating both sides with respect to their variables (noting that mass m is constant, so $ln\left(2m\right)$ differentiates to zero):

$$\frac{dK}{K}=2\frac{dp}{p}$$

Step 3: Compute the percentage increase in kinetic energy
Multiplying both sides of the differential relationship by 100 converts the fractional changes into percentage changes:

$$\left(\frac{\Delta K}{K}\times 100\right)=2\times \left(\frac{\Delta p}{p}\times 100\right)$$
We are given that the percentage increase in momentum is 0.01%:

$$\frac{\Delta p}{p}\times 100=0.01\%$$
Substituting this value into the percentage relation:

$$\text{Percentage increase in kinetic energy}=2\times 0.01\%=0.02\%$$
Thus, the kinetic energy of the body increases by 0.02%.