A metal tape is calibrated at 25^{circ} C. On | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The length of the metal tape at temperature $T$ is given by $L = L_0 [1 + \alpha \Delta T]$,where $L_0$ is the length at the calibration temperatu

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

(A) The length of the metal tape at temperature $T$ is given by $L = L_0 [1 + \alpha \Delta T]$,where $L_0$ is the length at the calibration temperature $T_0 = 25^{\circ} C$.The change in length is $\Delta L = L - L_0 = L_0 \alpha (T - T_0)$.The percentage error in the measurement is given by $|\frac{\Delta L}{L_0} 	imes 100\%|$.Substituting the given values: $\alpha = 1 	imes 10^{-5} {}^{\circ} C^{-1}$,$T = -15^{\circ} C$,and $T_0 = 25^{\circ} C$.Percentage error $= |\alpha (T - T_0) 	imes 100\%|$$= |1 	imes 10^{-5} (-15 - 25) 	imes 100\%|$$= |1 	imes 10^{-5} (-40) 	imes 100\%|$$= |-40 	imes 10^{-3}\%| = 0.04\%$.

$A$ metal tape is calibrated at $25^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the measurement of length is (Coefficient of linear expansion of metal $= 1 \times 10^{-5} {}^{\circ} C^{-1}$) (in $\%$)

Similar Questions

$A$ cylindrical metal rod of length $L_0$ is shaped into a ring with a small gap $x$ as shown. On heating the system:

If the volume of a block of metal changes by $0.12 \%$ when heated through $20^{\circ} C$,then find its coefficient of linear expansion.

Solids expand on heating because

$A$ bimetallic strip is made of aluminium and steel $(\alpha_{Al} > \alpha_{\text{steel}})$. On heating,the strip will

The length of a metal rod at $30^{\circ} C$ is $30 \ cm$. If its temperature is raised to $105^{\circ} C$, its length is increased by $0.027 \ cm$. Then the coefficient of linear expansion of the metal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module