JEE Advanced · Physics · 1. Math in Physics

In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)$ by using Searle's method, a wire of length $L=2 \mathrm{~m}$ and diameter $d=0.5 \mathrm{~mm}$ is used. For a load $M=2.5 \mathrm{~kg}$, an extension $l=0.25 \mathrm{~mm}$ in the length of the wire is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5 \mathrm{~mm}$. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the $Y$ measurement

A due to the errors in the measurements of $d$ and $l$ are the same.
B due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$.
C due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$.
D due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$.

Verified Solution

Answer & Solution

Correct Answer

(A) due to the errors in the measurements of $d$ and $l$ are the same.

Step-by-step Solution

Detailed explanation

The maximum possible error in $Y$ due to $l$ and $d$ $\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}$ Least count $=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$ $=\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}$

Here, $\Delta d=\Delta l=0.005 \mathrm{~mm}$

Error contribution of $l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}$

Error contribution of

$d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}$

Hence contribution to the maximum possible error in the measurement of $y$ due to $l$ and $d$ is the same.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Paragraph:
The figure shows a surface $X Y$ separating two transparent media, medium-1 and medium-2. The lines $a b$ and $c d$ represent wavefronts of a light wave travelling in medium-1 and incident on $X Y$. The lines ef and $g h$ represent wavefronts of the light wave in medium-2 after refraction.
Question:
The phases of the light wave at $c, d, e$ and $f$ are $\phi_c, \phi_d, \phi_e$ and $\phi_f$, respectively. It is given that $\phi_c \neq \phi_f$.

JEE Advanced 2007 Medium

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic process. Here

γ

is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.

Column – 1	Column – 2	Column – 3
(I) $W_{1 \rightarrow 2}=\frac{1}{\gamma-1}$ $\left(P_2 V_2-P_1 V_1\right)$	(i) Isothermal	(P)
(II) $W_{1 \rightarrow 2}=$ $-P V_2+P V_1$	(ii) Isochoric	(Q)
(III) $W_{1 \rightarrow 2}=0$	(iii) Isobaric	(R)
(IV) $W_{1 \rightarrow 2}=$ $-n R T \ln \left(\frac{V_2}{V_1}\right)$	(iv) Adiabatic	(S)

Which of the following options is the only correct representation of the process

∆ U = ∆ Q - P ∆ V

JEE Advanced 2017 Easy

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From JEE Advanced

let

f_{1} : R \to R, f_{2} : [0, \infty) \to R, f_{3} : R \to R a n d f_{4} : R \to [0, \infty)

be defined by

f_{1} (x) = \{\begin{matrix} |x| & i f & x < 0 \\ e^{x} & i f & x \geq 0 \end{matrix}; f_{2} (x) = x^{2}; f_{3} (x) = \{\begin{matrix} \sin x & i f & x < 0 \\ x & i f & x \geq 0 \end{matrix} a n d f_{4} (x) = \{\begin{matrix} f_{2} (f_{1} (x)) & i f & x < 0 \\ f_{2} (f_{1} (x)) - 1 & i f & x \geq 0 \end{matrix}

	List – I		List – II
(A)	$f_{4}$ is	(P)	Onto but not one – one
(B)	$f_{3}$ is	(Q)	neither continuous nor one-one
(C)	$f_{2}$ of $f_{1}$ is	(R)	differentiable but not one-one
(D)	$f_{2}$ is	(S)	continuous and one-one

JEE Advanced 2014 Medium

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 V$ . This potential drops to $0.6 V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\frac{h c}{e} = 1.24 \times 10^{- 6} J {mC}^{- 1}$ .]

JEE Advanced 2022 Easy

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