Question

A physical quantity is measured as q = (2.1 ± 0.5) units. calculate the percentage error in (1) q2 (2) 2q.

Answer 1

q2 it is 53%
2q is 38%..

Answer 2

Answer:

The percentage error in q²and 2 q are 23.81%.

Explanation:

Given that,

q= (2.1±0.5)

Let us consider,

(q±Δq)=(a±Δa)

Where, a = 2.1

Δa= 0.5

We know that, q = a

(1). The percentage error in q²

$(q\pm\Delta q)^2=(a\pm\Delta a)^2$

$q^2\pm2\Delta q\times q+(\Delta q)^2=a^2\pm2\Delta a\times a$

Dividing above equation by q²

$1\pm2\dfrac{\Delta q}{q}+(\dfrac{\Delta q}{q})^2=1\pm2\dfrac{\Delta a}{a}+(\dfrac{\Delta a}{a})^2$

On neglecting of higher power

$1\pm2\dfrac{\Delta q}{q}=1\pm2\dfrac{\Delta a}{a}$

$2\dfrac{\Delta q}{q}\times100=2\dfrac{\Delta a}{a}\times100$

$\dfrac{\Delta q}{q}\%=\pm\dfrac{0.5}{2.1}\times100$

$\dfrac{\Delta q}{q}\%=\pm23.81\%$

(2). The percentage error in 2 q

$2(q\pm\Delta q)=2(a\pm\Delta a)$

Dividing above equation by q

$1\pm\dfrac{\Delta q}{q}=1\pm\dfrac{\Delta a}{a}$

$\dfrac{\Delta q}{q}\times100=\dfrac{\Delta a}{a}\times100$

$\dfrac{\Delta q}{q}\%=\dfrac{0.5}{2.1}\times100$

$\dfrac{\Delta q}{q}\%=23.81\%$

Hence, The percentage error in q²and 2 q are 23.81%.