Q.27. The diameter of a wire measured by a screw gauge of least count cm is cm. The length measure

Q: What is the correct answer for JEE Main Physics Q27?

The correct answer is B. Sol. Young's modulus $Y = \frac{FL}{A\Delta L} = \frac{FL}{(\pi d^2/4)\Delta L} = \frac{4FL}{\pi d^2 \Delta L}$ Given values: Force $F = 100$ N (exact, so $\Delta F = 0$) Length $L = 150$ cm $= 1.5$ m. Least count $\Delta L_{scale} = 0.1$ cm $= 0.001$ m. Diameter $d = 0.08$ cm $= 0.0008$ m. Least count $\Delta d = 0.001$ cm $= 0.00001$ m. Extension $\Delta L = 0.5$ cm $= 0.005$ m. Least count $\D

Q: Q.27. The diameter of a wire measured by a screw gauge of least count cm is cm. The length measured by a scale of least count cm is cm. When a weight of N is applied to the wire, the extension…

The correct answer is B. Sol. Young's modulus $Y = \frac{FL}{A\Delta L} = \frac{FL}{(\pi d^2/4)\Delta L} = \frac{4FL}{\pi d^2 \Delta L}$ Given values: Force $F = 100$ N (exact, so $\Delta F = 0$) Length $L = 150$ cm $= 1.5$ m. Least count $\Delta L_{scale} = 0.1$ cm $= 0.001$ m. Diameter $d = 0.08$ cm $= 0.0008$ m. Least count $\Delta d = 0.001$ cm $= 0.00001$ m. Extension $\Delta L = 0.5$ cm $= 0.005$ m. Least count $\D

Q: What are the options for this JEE Main Physics question?

A. 3 · B. 65 · C. 13 · D. 25

Question

Q.27. The diameter of a wire measured by a screw gauge of least count  cm is  cm. The length measured by a scale of least count  cm is  cm. When a weight of  N is applied to the wire, the extension in length is  cm, measured by a micrometer of least count  cm. The error in the measured Young's modulus is . The value of  is ____.
(Ignore the contribution of the load to Young's modulus error calculation)

FirstInTest · Accepted Answer

Sol. Young's modulus $Y = \frac{FL}{A\Delta L} = \frac{FL}{(\pi d^2/4)\Delta L} = \frac{4FL}{\pi d^2 \Delta L}$

Given values:
Force $F = 100$ N (exact, so $\Delta F = 0$)
Length $L = 150$ cm $= 1.5$ m. Least count $\Delta L_{scale} = 0.1$ cm $= 0.001$ m.
Diameter $d = 0.08$ cm $= 0.0008$ m. Least count $\Delta d = 0.001$ cm $= 0.00001$ m.
Extension $\Delta L = 0.5$ cm $= 0.005$ m. Least count $\Delta (\Delta L) = 0.001$ cm $= 0.00001$ m.

The formula for fractional error in Y is:
$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta (\Delta L)}{\Delta L}$
Since $F$ is known exactly, $\Delta F = 0$.

$\frac{\Delta Y}{Y} = \frac{0.001}{1.5} + 2\frac{0.001}{0.08} + \frac{0.001}{0.5}$
$\frac{\Delta Y}{Y} = \frac{1}{1500} + 2\frac{1}{80} + \frac{1}{500}$
$\frac{\Delta Y}{Y} = 0.000666... + 0.025 + 0.002 = 0.027666... \approx 0.0277$

Now calculate $Y$:
$Y = \frac{4FL}{\pi d^2 \Delta L} = \frac{4 	imes 100 	imes 1.5}{\pi (0.08 	imes 10^{-2})^2 	imes 0.5 	imes 10^{-2}}$
$Y = \frac{600}{\pi (64 	imes 10^{-8}) (0.5 	imes 10^{-2})} = \frac{600}{\pi 	imes 32 	imes 10^{-10}} = \frac{600}{100.53 	imes 10^{-10}} \approx 5.968 	imes 10^{10} N/m^2$

Using the values from the solution:
$Y = \frac{100 	imes 1.5}{\frac{\pi (0.08)^2}{4} 	imes 0.5 	imes 10^{-2}} = \frac{100 	imes 1.5}{ (0.08)^2 	imes 0.5 	imes 10^{-2} 	imes \frac{\pi}{4} } = \frac{150}{0.0064 	imes 0.5 	imes 10^{-2} 	imes 0.785} = \frac{150}{0.002512 	imes 10^{-2}} \approx 5.97 	imes 10^{10} N/m^2$

Error in Young's modulus $\Delta Y = Y 	imes \frac{\Delta Y}{Y} = (5.97 	imes 10^{10}) 	imes 0.0277 \approx 1.65 	imes 10^9 N/m^2$.

So $\alpha = 1.65$.

Answer

3

Answer

65

Answer

13

Answer

25

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