A steel scale measures the length of a copper rod as L cm when both are at $20^{o} C$ , the calibration temperature for the scale. If the coefficients of linear expansion for steel and copper are $a_{s}$ and $a_{c}$ , respectively, what would be the scale reading (in cm) when both are at $21^{o} C$ ?

L \frac{(1 + a_{c})}{(1 + a_{s})}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

L \frac{a_{c}}{a_{s}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L \frac{a_{s}}{a_{c}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $L \frac{(1 + a_{c})}{(1 + a_{s})}$
The new length of the steel scale = $L (1 + α_{s})$
Thus the scale readings have shrunk by a factor of $1 + α_{s}$ .
The new length of the copper rod = $L (1 + α_{c})$ .
Thus the reading on the scale $= L (\frac{1 + α_{c}}{1 + α_{s}})$

Suggest Corrections

Similar questions

Q. A steel scale measures the length of a copper rod as

L cm

when both are at

20^{\circ} C

, the calibration temperature for the scale. If the coefficients of linear expansion for steel and copper are

α_{s}

and

α_{c}

respectively, what would be the scale reading (in cm) when both are at

21^{\circ} C

A steel scale measures the length of a copper rod as L cm when both are at $20^{\circ} C$ , the calibration temperature for the scale. If the coefficients of linear expansion for steel and copper are $α_{s}$ and $α_{c}$ respectively, what would be the scale reading (in cm) when both are at $21^{\circ} C$ ?