A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second.

The correct option is 7/5.

To find the ratio of the distance travelled by the body during the 4th and 3rd second, we can use the formula for the displacement of a body in the n^th second of its motion under uniform acceleration.
The distance travelled in the n^th second is given by:
$S_n=u+\frac{a}{2}(2n-1)$
where:
• $u$ is the initial velocity of the body,
• $a$ is the uniform acceleration,
• $n$ is the specific second of motion.

Since the body starts from rest, its initial velocity $u=0$.
Substituting $u=0$ into the formula, we get:
$S_n=\frac{a}{2}(2n-1)$

Now, let's calculate the distance travelled during the 4th second ($n=4$):
$S_4=\frac{a}{2}(2\times 4-1)=\frac{a}{2}(8-1)=\frac{7a}{2}$

Next, let's calculate the distance travelled during the 3rd second ($n=3$):
$S_3=\frac{a}{2}(2\times 3-1)=\frac{a}{2}(6-1)=\frac{5a}{2}$

Finally, we find the ratio of the distance travelled in the 4th second to that in the 3rd second:
$\frac{S_4}{S_3}=\frac{\frac{7a}{2}}{\frac{5a}{2}}=\frac{7}{5}$

Thus, the ratio of the distance travelled by the body during the 4th and 3rd second is 7/5.