If the velocity of body is doubled then % increase in centripetal force will be?

Since you are asking about centripetal force, I assume you are talking about a body moving in a circular path. In that case, speed is a better term than velocity. Or even better would be "instantaneous velocity", #v_i#. I say that because doubling velocity would imply it continues moving in a straight line.

The formula for centripetal force, #F_c# is

#F_c = (m*v_i^2)/r#

So, if #v_i# doubles, #F_c# would quadruple. The original #F_c# was #(m*v_i^2)/r# and it increased to #(4*m*v_i^2)/r#. That is an increase of #(3*m*v_i^2)/r# or 300%.

I hope this helps,
Steve

The centripetal force #F_c# on a body with mass #m# & linear velocity #v# moving on a circular path of radius #r# is given as

#F_c=\frac{mv^2}{r}#

Now, the centripetal force #F'_c# on the same body with mass #m# & with double linear velocity #2v# moving on the same circular path of radius #r# is given as

#F'_c=\frac{m(2v)^2}{r}=\frac{4mv^2}{r}#

Now, the percent #(%)# increase in the centripetal force

#\frac{F'_{c}-F_c}{F_c}\times 100#

#=\frac{\frac{4mv^2}{r}-\frac{mv^2}{r}}{\frac{mv^2}{r}}\times 100#

#=\frac{4-1}{1}\times 100#

#=300 #

hence the centripetal force #F_c# will increase by #300 %#