Question

90. With usual notation, the following equation, said to give the
distance covered in the nth second. i.e.,

S = u+a(2n-1)/2
(a) numerically correct only
(b) dimensionally correct only
(c)) both dimensionally and numerically only
(d) neither numerically nor dimensionally correct​

Answer 1

Answer:

c

Explanation:

The distance travelled in a time t is:

s=ut+12at2

So the distance travelled between t and t−Δt is:

Δs=s(t)−s(t−Δt)=ut+12at2−u(t−Δt)−12a(t−Δt)2=uΔt+12a(2tΔt−Δt2)

The equation you cite is obtained by setting Δt=1, but remember that you're setting Δt equal to one second not the dimensionless quantity 1. So your equation should really be:

Δs=u⋅(1 second)+12a(2t(1 second)−(1 second)2)

or multiplying this out:

Δs=u⋅(1 second)+at(1 second)−12a(1 second)2

So it is dimensionally consistent.

It is also numerically correct

Answer 2

The given equation is the equation for the distance covered by a body in nth second of its motion.

Sn =u + a/2 (2n−1)

Here, u is the initial velocity, a is the acceleration of the body and n is the second in which the distance is covered.

The dimension of Sn will be [LT−1] as it is the distance covered in 1 sec.

Substitute the dimensions in the equation.

[LT-1] = [LT−1] + [LT −2] [T]

[LT-1] = [LT−1] + [LT −2] [T][LT −1] = [LT−1]

Hence, the above equation is dimensionally & numerically correct.