JEE PYQs Integer Type
- To convert a physical quantity from one system of units to other system of units:
\(
n_2=n_1 \frac{u_1}{u_2}=n_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^a\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_2}\right]^b\left[\frac{\mathrm{~T}_1}{\mathrm{~T}_2}\right]^c
\) - Some of the commonly used systems of units are as follows
(i) CGS system: In this system, the units of length, mass and time are centimetre (cm), gram (g) and second (s), respectively. The unit of force in this system is dyne and that of work or energy is erg.
(ii) FPS system: In this system, the units of length, mass and time are foot (ft), pound (lb) and second (s), respectively. The unit of force in this system is poundal.
(iii) MKS system: In this system, the units of length, mass and time are metre (m), kilogram (kg) and second (s), respectively. The unit of force in this system is newton (N) and that of work or energy is joule (J).
(iv) International system (SI): This system of units helps in revolutionary changes over the MKS system and is known as rationalised MKS system. It is helpful to obtain all the physical quantities in physics. - Those physical quantities which are independent of each other and not defined in terms of other physical quantities, are called fundamental quantities or base quantities. The units of these quantities are called fundamental or base units.
- The quantities which can be expressed in terms of the fundamental quantities are called derived quantities. The units of these quantities are called derived units.
- Rules to determine significant figures
For determining number of significant figures, we use the following rules
Rule 1: All non-zero digits are significant, e.g. \(x=2567\) has four significant figures.
Rule 2: The zeros appearing between two non-zero digits are significant, no matter where the decimal point is, if at all, e.g. 6.028 has 4 significant figures.
Rule 3: If the number is less than 1 , the zero(s) on the right of decimal point but to the left of first non-zero digit are not significant.
e.g. 0.0042 has two significant digits.
Rule 4: The terminal or trailing zero(s) in a number without a decimal point are not significant. Thus, \(426 \mathrm{~m}=42600 \mathrm{~cm} =426000 \mathrm{~mm}\) has three significant figures.
Rule 5 In a number with decimal, zeros to the right of last non-zero digit are significant.
e.g. 4.600 or 0.002300 have four significant figures each. - Addition or subtraction:
Suppose in the measured values to be added or subtracted, the least number of significant digits after the decimal is \(n\). Then, in the sum or difference also, the number of significant digits after the decimal should be \(n\). - Multiplication or division:
Suppose in the measured values to be multiplied or divided, the least number of significant digits be \(n\), then in the product or quotient, the number of significant digits should also be \(n\).
Rules for rounding off a measurement: - Following are the rules for rounding off a measurement:
Rule 1: If the number lying to the right of cut-off digit is less than 5 , then the cut-off digit is retained as such. However, if it is more than 5 , then the cut-off digit is increased by 1. e.g. \(x=6.24\) is rounded off to 6.2 to two significant digits and \(x=5.328\) is rounded off to 5.33 to three significant digits.
Rule 2: If the insignificant digit to be dropped is 5 , then the rule is
(i) if the preceding digit is even, the insignificant digit is simply dropped.
(ii) if the preceding digit is odd, the preceding digit is raised by 1 .
e.g. \(x=6.265\) is rounded off to \(x=6.26\) to three significant digits and \(x=6.275\) is rounded off to \(x=6.28\) to three significant digits.
Rule 3: The exact numbers like \(\pi, 2,3\) and 4, etc., that appear in formulae and are known to have infinite significant figures, can be rounded off to a limited number of significant figures as per the requirement. - Error \(=\) True value – Measured value
- Absolute error: The difference between the true value and the measured value of a quantity is called an absolute error. Usually the mean value \(a_m\) is taken as the true value. So, if
\(
a_m=\frac{a_1+a_2+\ldots+a_n}{n}=\frac{1}{n} \sum_{i=1}^n a_i
\)
Then by definition, absolute errors in the measured values of the quantity are
\(
\begin{gathered}
\Delta a_1=a_1-a_m \\
\Delta a_2=a_2-a_m \\
\vdots \\
\vdots \\
\Delta a_n=a_n-a_m
\end{gathered}
\)
Absolute error may be positive or negative.
Mean absolute error It is the arithmetic mean of the magnitudes of absolute errors. Thus,
\(
\Delta a_{\text {mean }}=\frac{\left|\Delta a_1\right|+\left|\Delta a_2\right|+\ldots+\left|\Delta a_n\right|}{n}=\frac{1}{n} \sum_{i=1}^n\left|\Delta a_i\right|
\)
Thus final result of measurement can be written as
\(
a=a_m \pm \Delta a_{\text {mean }}
\)
This implies that value of \(a\) is likely to lie between \(a_m+\Delta a_{\text {mean }}\) and \(a_m-\Delta a_{\text {mean }}\). - Relative or fractional error: The ratio of mean absolute error to the mean value of the quantity measured is called relative or fractional error.
Thus, \(\quad\) Relative error \(=\frac{\Delta a_{\text {mean }}}{a_m}\)
Percentage error When the relative error is expressed in percent, it is called percentage error. It is denoted by \(\delta a\).
Thus, \(\quad \delta a=\frac{\Delta a_{\text {mean }}}{a_m} \times 100 \%\) - Error in sum or difference: Let \(x=a \pm b\):
\(
\Delta x= \pm(\Delta a+\Delta b)
\) - Error in product: Let \(x=a b\):
\(
\frac{\Delta x}{x}= \pm\left(\frac{\Delta a}{a}+\frac{\Delta b}{b}\right)
\) - Error in division: Let \(x=\frac{a}{b}\):
\(
\frac{\Delta x}{x}= \pm\left(\frac{\Delta a}{a}+\frac{\Delta b}{b}\right)
\) - Error in quantity raised to some power:
Let \(x=\frac{a^n}{b^m}\)
\(
\frac{\Delta x}{x}= \pm\left(n \frac{\Delta a}{a}+m \frac{\Delta b}{b}\right)
\)
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- Question 1 of 20
1. Question
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of \(\left[M^a L^b T^c\right]\). If \(b=-3\), the value of \(c\) is ____. [JEE Main 2025 January]
CorrectIncorrectHint
(a) Step 1: Determine the dimensions of Modulus of Elasticity
Modulus of elasticity has the same dimensions as stress (force per unit area).
\(
\text { Dimensions of Modulus }=\frac{\text { Force }}{\text { Area }}=\frac{\left[M L T^{-2}\right]}{\left[L^2\right]}=\left[M L^{-1} T^{-2}\right]
\)
Step 2: Determine the dimensions of Torque
Torque is defined as force multiplied by a perpendicular distance.
\(
\text { Dimensions of Torque }=\text { Force × Distance }=\left[M L T^{-2}\right] \times[L]=\left[M L^2 T^{-2}\right]
\)
Step 3: Determine the dimensions of the final quantity
The required quantity is “modulus of elasticity per unit torque applied on the system”.
\(
\begin{aligned}
& \text { Dimensions of Quantity }=\frac{\text { Dimensions of Modulus }}{\text { Dimensions of Torque }}=\frac{\left[M L^{-1} T^{-2}\right]}{\left[M L^2 T^{-2}\right]} \\
& \text { Dimensions of Quantity }=\left[M^{1-1} L^{-1-2} T^{-2-(-2)}\right]=\left[M^0 L^{-3} T^0\right]
\end{aligned}
\)
Step 4: Compare with the given dimensional formula
The derived dimensions \(\left[\boldsymbol{M}^0 \boldsymbol{L}^{-3} \boldsymbol{T}^0\right]\) are compared to the given form \(\left[\boldsymbol{M}^a \boldsymbol{L}^b \boldsymbol{T}^c\right]\).
By comparison: \(a=0, b=-3\), and \(c=0\). - Question 2 of 20
2. Question
A physical quantity \(C\) is related to four other quantities \(p, q, r\) and \(s\) as follows
\(
C=\frac{p q^2}{r^3 \sqrt{s}}
\)
The percentage errors in the measurement of \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) and s are \(1 \%, 2 \%, 3 \%\) and \(2 \%\), respectively. The percentage error in the measurement of \(C\) will be ____ % [JEE Main 2025 (Online) 3rd April Evening Shift]CorrectIncorrectHint
(c) To determine the percentage error in the measurement of \(C\), which is related to \(p, q, r\) and \(s\) as:
\(
C=\frac{p q^2}{r^3 \sqrt{s}}
\)
we first express it in terms of powers:
\(
C=p^1 q^2 r^{-3} s^{-1 / 2}
\)
The percentage error in \(C\) can be calculated using the formula for the propagation of error, which is:
\(
\left(\frac{\Delta C}{C}\right)_{\max }=\left|\frac{\Delta p}{p}\right|+2\left|\frac{\Delta q}{q}\right|+3\left|\frac{\Delta r}{r}\right|+\frac{1}{2}\left|\frac{\Delta s}{s}\right|
\)
Given the percentage errors for \(p, q, r\) and \(s\) are \(1 \%, 2 \%, 3 \%\), and \(2 \%\) respectively, we substitute these values into the formula:
\(
\left(\frac{\Delta C}{C}\right)_{\max }=1 \%+2 \times 2 \%+3 \times 3 \%+\frac{1}{2} \times 2 \%
\)
Calculating each part:
Contribution from \(p: 1 \%\)
Contribution from \(q\) : \(4 \%\) (since \(2 \times 2 \%=4 \%\) )
Contribution from \(r: 9 \%\) (since \(3 \times 3 \%=9 \%\) )
Contribution from \(s: 1 \%\) (since \(\frac{1}{2} \times 2 \%=1 \%\) )
Adding these contributions together:
\(
1 \%+4 \%+9 \%+1 \%=15 \%
\)
Thus, the maximum percentage error in the measurement of \(C\) is \(15 \%\). - Question 3 of 20
3. Question
A physical quantity \(Q\) is related to four observables \(a, b, c, d\) as follows:
\(
Q=\frac{a b^4}{c d}
\)
where, \(\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2}\) and \(\mathrm{d}=(50 \pm 0.1) \mathrm{m}\), then the percentage error in \(Q\) is \(\frac{x}{1000}\), where \(x=\) _____ . [JEE Main 2025 (Online) 29th January Evening Shift]CorrectIncorrectHint
(a) Given, \(Q=\frac{a b^4}{c d}\)
\(
\begin{aligned}
& a=(60 \pm 3) P a \Rightarrow a=60 P a, \Delta a=3 P a \\
& b=(20 \pm 0.1) m \Rightarrow b=20 m, \Delta b=0.1 m \\
& c=(40 \pm 0.2) N s m^{-2} \Rightarrow c=40 N s m^{-2}, \Delta c=0.2 N s m^{-2} \\
& d=(50 \pm 0.1) m \Rightarrow d=50 m, \Delta d=0.1 m
\end{aligned}
\)
As, \(Q=\frac{a b^4}{c d}\)
by taking \(\ln\) on both sides,
\(
\ln Q=\ln a+u \ln b-\ln c-\ln d
\)
Now, by differentiating,
\(
\frac{d Q}{Q}=\frac{d a}{a}+4 \frac{d b}{b}-\frac{d c}{c}-\frac{d d}{d}
\)
So, maximum fractional error in \(Q\) is given by,
\(
\begin{aligned}
& \frac{\Delta Q}{Q}=\frac{\Delta a}{a}+4 \frac{\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d}{d} \\
& \Rightarrow \frac{\Delta Q}{Q}=\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\frac{0.2}{40}+\frac{0.1}{40} \\
& =\frac{1}{20}+\frac{1}{50}+\frac{1}{200}+\frac{1}{500} \\
& \Rightarrow \frac{\Delta Q}{Q}=\frac{50+20+5+2}{1000}=\frac{77}{1000}
\end{aligned}
\)
Hence the \% error in \(Q=\frac{\Delta Q}{Q} \times 100 \%\)
\(
=\frac{7700}{1000} \%=\frac{x}{1000} \text { (given) }
\)
So, \(x=7700\) - Question 4 of 20
4. Question
A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be \(\frac{x}{100}\) where \(x\) is _____. [JEE Main 2025 (Online) 28th January Morning Shift]
CorrectIncorrectHint
(c) Step 1: Calculate the least count
The least count (LC) of the screw gauge is determined by the ratio of the pitch to the number of divisions on the circular scale.
\(
\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}
\)
Given Pitch \(=0.75 \mathrm{~mm}\) and number of divisions \(=15\), the least count is:
\(
\mathrm{LC}=\frac{0.75 \mathrm{~mm}}{15}=0.05 \mathrm{~mm}
\)
Step 2: Determine the fractional error formula
The area of a rectangular sheet is calculated as \(\boldsymbol{A}=\boldsymbol{L} \times \boldsymbol{B}\). The maximum fractional error in the area ( \(\frac{\Delta A}{A}\) ) is the sum of the fractional errors in the length and breadth:
\(
\frac{\Delta A}{A}=\frac{\Delta L}{L}+\frac{\Delta B}{B}
\)
For a measurement using a screw gauge, the maximum absolute error ( \(\boldsymbol{\Delta} \boldsymbol{L}\) or \(\boldsymbol{\Delta} \boldsymbol{B}\) ) is equal to the least count (LC).
Step 3: Calculate the maximum fractional error
Substitute the values into the formula:
\(
\begin{gathered}
\frac{\Delta A}{A}=\frac{0.05 \mathrm{~mm}}{5 \mathrm{~mm}}+\frac{0.05 \mathrm{~mm}}{2.5 \mathrm{~mm}} \\
\frac{\Delta A}{A}=0.01+0.02 \\
\frac{\Delta A}{A}=0.03
\end{gathered}
\)
Step 4: Determine the value of \(x\)
The problem asks for the maximum fractional error to be expressed in the form \(\frac{x}{100}\).
\(
\frac{\Delta A}{A}=0.03=\frac{3}{100}
\)
Comparing the expressions, the value of \(x\) is 3. - Question 5 of 20
5. Question
The least count of a screw guage is 0.01 mm . If the pitch is increased by \(75 \%\) and number of divisions on the circular scale is reduced by \(50 \%\), the new least count will be ____ \(\times 10^{-3} \mathrm{~mm}\) [JEE Main 2025 (Online) 24th January Morning Shift]
CorrectIncorrectHint
(d) Step 1: Define initial parameters and formula
The initial least count ( \(L C_1\) ) is given as 0.01 mm. The formula for the least count is the ratio of pitch \((P)\) to the number of divisions on the circular scale \((N)\) :
\(
L C=\frac{P}{N}
\)
So, initially, \(L C_1=\frac{P_1}{N_1}=0.01 \mathrm{~mm}\).
Step 2: Calculate new pitch and number of divisions
The pitch is increased by \(75 \%\), resulting in a new pitch \(\left(\boldsymbol{P}_{\mathbf{2}}\right)\) :
\(
P_2=P_1+0.75 P_1=1.75 P_1
\)
The number of divisions on the circular scale is reduced by \(50 \%\), resulting in a new number of divisions ( \(\boldsymbol{N}_{\mathbf{2}}\) ):
\(
N_2=N_1-0.50 N_1=0.50 N_1
\)
Step 3: Calculate the new least count
The new least count ( \(L C_2\) ) is calculated using the new parameters \(P_2\) and \(N_2\) :
\(
L C_2=\frac{P_2}{N_2}=\frac{1.75 P_1}{0.50 N_1}=3.5 \times \frac{P_1}{N_1}
\)
Substitute the value of \(L C_1\) into the equation:
\(
L C_2=3.5 \times 0.01 \mathrm{~mm}=0.035 \mathrm{~mm}=35 \times 10^{-3} \mathrm{~mm}
\) - Question 6 of 20
6. Question
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while \(46^{\text {th }}\) division the circular scale coincide with the reference line. The diameter of the wire is ____ \(\times 10^{-2} \mathrm{~mm}\). [JEE Main 2023 (Online) 30th January Morning Shift]
CorrectIncorrectHint
(b) Step 1: Calculate the least count
The least count (LC) of the screw gauge is calculated using the formula:
\(
\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}
\)
Given pitch \((P)=0.5 \mathrm{~mm}\) and number of divisions on circular scale \((N)=100\).
\(
\mathrm{LC}=\frac{0.5 \mathrm{~mm}}{100}=0.005 \mathrm{~mm}
\)
Step 2: Determine the zero error
The zero of the circular scale is 6 divisions below the reference line when the studs are in contact, which indicates a positive zero error.
\(
\text { Zero Error }=+ \text { (Number of divisions below the line) } \text { × } \text { LC }
\)
\(
\text { Zero Error }=6 \times 0.005 \mathrm{~mm}=+0.03 \mathrm{~mm}
\)
The zero correction (ZC) is the negative of the zero error:
\(
\text { Zero Correction }=-0.03 \mathrm{~mm}
\)
Step 3: Calculate the observed reading
When the wire is placed between the studs, the main scale reading (MSR) is determined by the number of visible linear scale divisions. Each linear scale division is 0.5 mm (equal to the pitch).
\(
\mathrm{MSR}=4 \times 0.5 \mathrm{~mm}=2.0 \mathrm{~mm}
\)
The circular scale reading (CSR) is the division that coincides with the reference line:
\(
\mathrm{CSR}=46 \times \mathrm{LC}=46 \times 0.005 \mathrm{~mm}=0.23 \mathrm{~mm}
\)
The observed reading (measured thickness) is the sum of MSR and CSR:
\(
\text { Observed Reading = MSR + CSR = } 2.0 \mathrm{~mm}+0.23 \mathrm{~mm}=2.23 \mathrm{~mm}
\)
Step 4: Calculate the actual diameter
The actual diameter is the observed reading minus the zero error (or plus the zero correction):
\(
\begin{aligned}
& \text { Actual Diameter }=\text { Observed Reading }- \text { Zero Error } \\
& \text { Actual Diameter }=2.23 \mathrm{~mm}-0.03 \mathrm{~mm}=2.20 \mathrm{~mm}
\end{aligned}
\)
Step 5: Convert to required format
The question asks for the diameter in the format \(\times 10^{-2} \mathrm{~mm}\).
\(
\text { Actual Diameter }=2.20 \mathrm{~mm}=220 \times 10^{-2} \mathrm{~mm}
\) - Question 7 of 20
7. Question
In an experiment to find acceleration due to gravity ( g ) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, The accuracy in the determination of g is found to be \(x \%\). The value of \(x\) is _____ . [JEE Main 2022 (Online) 28th July Evening Shift]
CorrectIncorrectHint
(d)
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& \frac{d g}{g} \times 100=\frac{2 d T}{T} \times 100+\frac{d l}{l} \times 100 \\
& =2 \times \frac{1}{50} \times 100+\frac{1}{100} \times 100=5 \%
\end{aligned}
\) - Question 8 of 20
8. Question
The one division of main scale of Vernier callipers reads 1 mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2 cm and \(6^{\text {th }}\) Vernier division coincides scale division. The diameter of the bob will be ____ \(\times 10^{-2} \mathrm{~cm}\). [JEE Main 2022 (Online) 27th July Morning Shift]
CorrectIncorrectHint
(b) Least Count (LC):
1 main scale division (MSD) \(=1 \mathrm{~mm}=0.1 \mathrm{~cm}\).
10 vernier scale divisions (VSD) \(=9\) MSD. So, 1 VSD \(=0.9\) MSD \(=0.9 \mathrm{~mm}=0.09\) cm.
\(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.1 \mathrm{~cm}-0.09 \mathrm{~cm}=0.01 \mathrm{~cm}\).
Zero Error:
Zero of the Vernier scale is to the right of the main scale zero (positive zero error), and the 4th VSD coincides with a main scale division.
Zero Error (ZE) = Coinciding VSD number × LC = \(4 \times 0.01 \mathrm{~cm}=+0.04 \mathrm{~cm}\).
Observed Reading:
Main Scale Reading (MSR) is the reading just before the zero mark of the Vernier scale, which is 4.1 cm.
Vernier Scale Reading (VSR) = Coinciding VSD number × LC = \(6 \times 0.01 \mathrm{~cm}=0.06\) cm.
Observed Reading \((\mathrm{OR})=\mathrm{MSR}+\mathrm{VSR}=4.1 \mathrm{~cm}+0.06 \mathrm{~cm}=4.16 \mathrm{~cm}\).
Corrected Reading (Diameter):
Actual Reading = Observed Reading – Zero Error.
Actual Reading \(=4.16 \mathrm{~cm}-(+0.04 \mathrm{~cm})=4.12 \mathrm{~cm}\).
The diameter of the bob is 4.12 cm , which is \(412 \times 10^{-2} \mathrm{~cm}\). - Question 9 of 20
9. Question
The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of \((-0.05) \mathrm{cm}\). While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be _____ \(\times 10^{-2} \mathrm{~cm}\). [JEE Main 2022 (Online) 29th June Evening Shift]
CorrectIncorrectHint
(a) Step 1: Calculate Observed Reading
The observed reading (OR) is the sum of the main scale reading (MSR) and the product of the coinciding vernier division (CVD) and the least count (LC).
The least count in cm is \(0.1 \mathrm{~mm}=0.01 \mathrm{~cm}\).
The formula for the observed reading is \(\mathrm{OR}=\mathrm{MSR}+(\mathrm{CVD} \times \mathrm{LC})\).
\(
\begin{gathered}
\mathrm{OR}=1.7 \mathrm{~cm}+(5 \times 0.01 \mathrm{~cm}) \\
\mathrm{OR}=1.7 \mathrm{~cm}+0.05 \mathrm{~cm}=1.75 \mathrm{~cm}
\end{gathered}
\)
Step 2: Calculate Corrected Reading
The corrected reading (CR) is obtained by subtracting the zero error (ZE) from the observed reading. The zero error is given as -0.05 cm.
The formula for the corrected reading is \(\mathrm{CR}=\mathrm{OR}-\mathrm{ZE}\).
\(
\begin{gathered}
\mathrm{CR}=1.75 \mathrm{~cm}-(-0.05 \mathrm{~cm}) \\
\mathrm{CR}=1.75 \mathrm{~cm}+0.05 \mathrm{~cm}=1.80 \mathrm{~cm}
\end{gathered}
\)
We need the value in the form \(\times 10^{-2} \mathrm{~cm}\).
\(
1.80 \mathrm{~cm}=180 \times 10^{-2} \mathrm{~cm}
\) - Question 10 of 20
10. Question
A student in the laboratory measures thickness of a wire using screw gauge. The readings are \(1.22 \mathrm{~mm}, 1.23 \mathrm{~mm}, 1.19 \mathrm{~mm}\) and 1.20 mm. The percentage error is \(\frac{x}{121} \%\). The value of \(x\) is _____. [JEE Main 2022 (Online) 28th June Evening Shift]
CorrectIncorrectHint
(d)
\(
\begin{aligned}
& I_{\text {mean }}=\frac{1.22+1.23+1.19+1.20}{4}=1.21 \\
& \Delta I_{\text {mean }}=\frac{0.01+0.02+0.02+0.01}{4}=0.015 \\
& \text { So } \% I=\frac{\Delta I_{\text {mean }}}{I_{\text {mean }}} \times 100=\frac{0.015}{1.21} \times 100=\frac{150}{121} \% \\
& x=150
\end{aligned}
\) - Question 11 of 20
11. Question
A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is _____ \(\times 10^{-6} \mathrm{~m}\). [JEE Main 2022 (Online) 26th June Evening Shift]
CorrectIncorrectHint
(d) Step 1: Determine the value of one main scale division (MSD)
The main scale has 40 divisions in 1 cm , which is 0.01 m . The value of one MSD in meters is calculated as:
\(
1 \mathrm{MSD}=\frac{0.01 \mathrm{~m}}{40}=0.00025 \mathrm{~m}
\)
In the required format, this is \(250 \times 10^{-6} \mathrm{~m}\).
Step 2: Determine the relationship between MSD and VSD
It is given that 50 Vernier scale divisions (VSD) are equal to 49 main scale divisions (MSD). This allows for the calculation of the value of one VSD:
\(
\begin{gathered}
50 \mathrm{VSD}=49 \mathrm{MSD} \\
1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}=\frac{49}{50} \times 0.00025 \mathrm{~m}=0.000245 \mathrm{~m}
\end{gathered}
\)
Step 3: Calculate the least count (LC)
The least count of the Vernier scale system is the difference between one main scale division and one Vernier scale division:
\(
\begin{gathered}
\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\
\mathrm{LC}=0.00025 \mathrm{~m}-0.000245 \mathrm{~m}=0.000005 \mathrm{~m}
\end{gathered}
\)
Expressing this value in the required format \(\times 10^{-6} \mathrm{~m}\) :
\(
\mathrm{LC}=5 \times 10^{-6} \mathrm{~m}
\) - Question 12 of 20
12. Question
In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be ______ \(\times 10^{-2} \mathrm{~mm}\). [JEE Main 2022 (Online) 26th June Morning Shift]
CorrectIncorrectHint
(a) Step 1: Determine main scale division length
The main scale division (MSD) length is defined as \(1 \mathrm{MSD}=\frac{1}{20} \mathrm{~cm}\). In millimeters, this is \(1 \mathrm{MSD}=0.5 \mathrm{~mm}\).
Step 2: Calculate one Vernier scale division length
The relationship between the Vernier scale divisions (VSD) and main scale divisions is \(10 \mathrm{VSD}=9 \mathrm{MSD}\). Therefore, one VSD is calculated as
\(
1 \mathrm{VSD}=\frac{9}{10} \times 0.5 \mathrm{~mm}=0.45 \mathrm{~mm} .
\)
Step 3: Calculate the least count
The least count (LC) is the difference between one main scale division and one Vernier scale division: \(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=(0.50-0.45) \mathrm{mm}\).
Step 4: Finalize the least count value
The final value of the least count is calculated as \(\mathrm{LC}=0.05 \mathrm{~mm}\), which can also be expressed as \(\mathrm{LC}=5 \times 10^{-2} \mathrm{~mm}\). - Question 13 of 20
13. Question
For \(z=a^2 x^3 y^{\frac{1}{2}}\), where ‘ \(a\) ‘ is a constant. If percentage error in measurement of ‘ \(x\) ‘ and ‘ \(y\) ‘ are \(4 \%\) and \(12 \%\) respectively, then the percentage error for ‘ \(z\) ‘ will be _____ %. [JEE Main 2022 (Online) 25th June Evening Shift]
CorrectIncorrectHint
(c) The maximum fractional error in a quantity \(\mathbf{z}\) defined as \(\mathbf{z}=k x^m y^n\), where \(k\) is a constant, is given by the formula:
\(
\frac{\Delta z}{z}=\left(m \frac{\Delta x}{x}+n \frac{\Delta y}{y}\right)
\)
The corresponding percentage error is:
\(
\% \Delta \mathrm{z}=(m \times \% \Delta x+n \times \% \Delta y)
\)
From the given equation \(z=a^2 x^3 y^{\frac{1}{2}}\), we have:
Power of \(x, m=3\)
Power of \(y, n=\frac{1}{2}\)
Percentage error in \(x, \% \Delta x=4 \%\)
Percentage error in \(y, \% \Delta y=12 \%\)
Substitute these values into the percentage error formula:
\(
\begin{gathered}
\% \Delta z=\left(3 \times 4 \%+\frac{1}{2} \times 12 \%\right) \\
\% \Delta z=(12 \%+6 \%) \\
\% \Delta z=18 \%
\end{gathered}
\) - Question 14 of 20
14. Question
The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and \(8^{\text {th }}\) division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is ____ \(\times 10^{-2} \mathrm{~cm}\). [JEE Main 2021 (Online) 31st August Evening Shift]
CorrectIncorrectHint
(b) Step 1: Determine the Least Count
First, calculate the least count (LC) of the vernier callipers.
The Main Scale Division (MSD) is 1 mm or 0.1 cm.
Given that 9 divisions of the main scale equal 10 divisions of the vernier scale (VSD), we find the value of one VSD as:
\(
1 \mathrm{VSD}=\frac{9}{10} \times \mathrm{MSD}=0.9 \mathrm{~mm}=0.09 \mathrm{~cm}
\)
The least count is the difference between one MSD and one VSD:
\(
\mathrm{LC}=\mathrm{MSD}-\mathrm{VSD}=1 \mathrm{~mm}-0.9 \mathrm{~mm}=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}
\)
Step 2: Calculate the Observed Diameter
Next, determine the observed diameter using the main scale reading (MSR) and the vernier scale coincidence (VSC):
\(
\text { Observed Reading }=\mathrm{MSR}+(\mathrm{VSC} \times \mathrm{LC})
\)
Given MSR is \(10 \mathrm{~mm}(1.0 \mathrm{~cm})\) and VSC is the \(8^{\text {th }}\) division:
\(
\text { Observed Reading }=1.0 \mathrm{~cm}+(8 \times 0.01 \mathrm{~cm})=1.0 \mathrm{~cm}+0.08 \mathrm{~cm}=1.08 \mathrm{~cm}
\)
Step 3: Apply the Zero Correction
The vernier callipers have a positive zero error of 0.04 cm. The true diameter is calculated by subtracting the zero error from the observed reading:
\(
\begin{aligned}
& \text { True } \text { Diameter }=\text { Observed Reading }- \text { Zero Error } \\
& \text { True Diameter }=1.08 \mathrm{~cm}-0.04 \mathrm{~cm}=1.04 \mathrm{~cm}
\end{aligned}
\)
Step 4: Calculate the Radius
The problem asks for the radius of the spherical bob, which is half of the diameter:
\(
\begin{aligned}
\text { Radius } & =\frac{\text { True Diameter }}{2} \\
\text { Radius } & =\frac{1.04 \mathrm{~cm}}{2}=0.52 \mathrm{~cm}
\end{aligned}
\)
To express the radius in the required format \(\left(\times 10^{-2} \mathrm{~cm}\right)\) :
\(
\text { Radius }=52 \times 10^{-2} \mathrm{~cm}
\) - Question 15 of 20
15. Question
The acceleration due to gravity is found upto an accuracy of \(4 \%\) on a planet. The energy supplied to a simple pendulum to known mass ‘ \(m\) ‘ to undertake oscillations of time period \(T\) is being estimated. If time period is measured to an accuracy of \(3 \%\), the accuracy to which \(E\) is known as _____ % [JEE Main 2021 (Online) 26th August Evening Shift]
CorrectIncorrectHint
(c)
\(
\begin{aligned}
&T=2 \pi \sqrt{\frac{l}{g}} \Rightarrow l=\frac{T^2 g}{4 \pi^2}\\
&E=m g l \frac{\theta^2}{2}=m g^2 \frac{T^2 \theta^2}{8 \pi^2}\\
&\frac{d E}{E}=2\left(\frac{d g}{g}+\frac{d T}{T}\right)\\
&=(4+3)=14 \%
\end{aligned}
\) - Question 16 of 20
16. Question
Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students \(A\) and \(B\) is _____. [JEE Main 2021 (Online) 25th July Morning Shift]
[Figure shows position of reference ‘ 0 ‘ when jaws of screw gauge are closed]
Given pitch \(=0.1 \mathrm{~cm}\).
CorrectIncorrectHint
(c) Step 1: Determine the least count and zero errors
The least count (LC) of the screw gauge is calculated using the given pitch and number of divisions:
\(
\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of circular divisions }}=\frac{0.1 \mathrm{~cm}}{100}=0.001 \mathrm{~cm}
\)
From the problem’s implied figures and associated solutions, Screw gauge (A) has a positive zero error of +5 divisions, and Screw gauge (B) has a negative zero error of -8 divisions.
Zero Error \((\mathrm{A})=+5 \times \mathrm{LC}=+5 \times 0.001 \mathrm{~cm}=+0.005 \mathrm{~cm}\)
Zero Error \((\mathrm{B})=-8 \times \mathrm{LC}=-8 \times 0.001 \mathrm{~cm}=-0.008 \mathrm{~cm}\)
Step 2: Determine the observed readings
The actual value of the radius is 0.322 cm. The corrected reading is given by the formula:
Corrected Reading = Observed Reading – Zero Error
Therefore, the observed reading is:
Observed Reading = Corrected Reading + Zero Error
Observed Reading \((\mathrm{A})=0.322 \mathrm{~cm}+0.005 \mathrm{~cm}=0.327 \mathrm{~cm}\)
Observed Reading \((\mathrm{B})=0.322 \mathrm{~cm}+(-0.008 \mathrm{~cm})=0.314 \mathrm{~cm}\)
The observed reading is the sum of the main scale reading (MSR) and the circular scale reading (CSR) value:
Observed Reading = MSR + CSR value
Assuming standard MSR values from the context of similar problems:
For (A), the MSR is 0.3 cm . The CSR value is \(0.327 \mathrm{~cm}-0.3 \mathrm{~cm}=0.027 \mathrm{~cm}\).
For (B), the MSR is 0.3 cm (or possibly 0.2 cm in some interpretations, but we will use the determined value). Using the 0.314 cm observed reading, the CSR value is \(0.314 \mathrm{~cm}-0.3 \mathrm{~cm}=0.014 \mathrm{~cm}\).
Step 3: Calculate the difference in circular scale readings (divisions)
The circular scale reading (CSR) in divisions is the CSR value divided by the least count.
\(\operatorname{CSR}(\mathrm{A})\) in divisions \(=\frac{0.027 \mathrm{~cm}}{0.001 \mathrm{~cm}}=27\) divisions
\(\operatorname{CSR}(B)\) in divisions \(=\frac{0.014 \mathrm{~cm}}{0.001 \mathrm{~cm}}=14\) divisions
The absolute value of the difference between the final circular scale readings observed by the students in divisions is:
\(
|\operatorname{CSR}(\mathrm{A})-\operatorname{CSR}(\mathrm{B})|=|27-14|=13
\) - Question 17 of 20
17. Question
Three students \(S_1, S_2\) and \(S_3\) perform an experiment for determining the acceleration due to gravity ( g ) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table. [JEE Main 2021 (Online) 22th July Evening Shift]
\(
\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Student } \\
\text { No. }
\end{array} & \begin{array}{c}
\text { Length of } \\
\text { pendulum } \\
(\mathrm{cm})
\end{array} & \begin{array}{c}
\text { No. of } \\
\text { oscillations } \\
(\mathrm{n})
\end{array} & \begin{array}{c}
\text { Total time for } \\
\text { n oscillations }
\end{array} & \begin{array}{c}
\text { Time } \\
\text { period } \\
(\mathrm{s})
\end{array} \\
\hline 1 . & 64.0 & 8 & 128.0 & 16.0 \\
\hline 2 . & 64.0 & 4 & 64.0 & 16.0 \\
\hline 3 . & 20.0 & 4 & 36.0 & 9.0 \\
\hline
\end{array}
\)
(Least count of length \(=0.1 \mathrm{~cm}\) and Least count for time \(=0.1 \mathrm{~s}\) )
If \(\mathrm{E}_1, \mathrm{E}_2\) and \(\mathrm{E}_3\) are the percentage errors in ‘ g ‘ for students 1,2 and 3 respectively, then the minimum percentage error is obtained by student no. _____. [JEE Main 2021 (Online) 22th July Evening Shift]CorrectIncorrectHint
(a) Step 1: Determine the formula for percentage error
The formula for acceleration due to gravity (g) using a simple pendulum is \(g=\frac{4 \pi^2 L}{T^2}\), where \(L\) is the length and \(T\) is the time period. The fractional error in \(g\) is given by the error propagation formula:
\(
\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \frac{\Delta T}{T}
\)
where \(\boldsymbol{\Delta} \boldsymbol{L}\) is the absolute error in length \((0.1 \mathrm{~cm})\) and \(\boldsymbol{\Delta} \boldsymbol{T}\) is the absolute error in the time period. The time period \(T=\frac{t}{n}\), so \(\Delta T=\frac{\Delta t_0}{n}\), where \(\Delta t_0\) is the least count of the stopwatch ( 0.1 s ) and \(n\) is the number of oscillations.
Step 2: Calculate the percentage error for each student The percentage error
\(
E=\frac{\Delta g}{g} \times 100=\frac{\Delta l}{l} \times 100+2 \frac{\Delta T}{T} \times 100
\)
For student 1: \(L_1=64.0 \mathrm{~cm}, n_1=8, t_1=128.0 \mathrm{~s}, T_1=16.0 \mathrm{~s}\).
\(
\begin{aligned}
& E_1=\left(\frac{\Delta L}{L_1}+2 \frac{\Delta T_1}{T_1}\right) \times 100=\left(\frac{0.1}{64.0}+2 \frac{0.1 / 8}{16.0}\right) \times 100 \\
& E_1=\left(\frac{0.1}{64.0}+\frac{0.2}{128.0}\right) \times 100=\left(\frac{0.1}{64.0}+\frac{0.1}{64.0}\right) \times 100=\frac{0.2}{64.0} \times 100 \approx 0.3125 \%
\end{aligned}
\)
For student 2: \(L_2=64.0 \mathrm{~cm}, n_2=4, t_2=64.0 \mathrm{~s}, T_2=16.0 \mathrm{~s}\).
\(
\begin{aligned}
& E_2=\left(\frac{\Delta L}{L_2}+2 \frac{\Delta T_2}{T_2}\right) \times 100=\left(\frac{0.1}{64.0}+2 \frac{0.1 / 4}{16.0}\right) \times 100 \\
& E_2=\left(\frac{0.1}{64.0}+\frac{0.2}{64.0}\right) \times 100=\frac{0.3}{64.0} \times 100 \approx 0.46875 \%
\end{aligned}
\)
For student 3: \(L_3=20.0 \mathrm{~cm}, n_3=4, t_3=36.0 \mathrm{~s}, T_3=9.0 \mathrm{~s}\).
\(
\begin{aligned}
& E_3=\left(\frac{\Delta L}{L_3}+2 \frac{\Delta T_3}{T_3}\right) \times 100=\left(\frac{0.1}{20.0}+2 \frac{0.1 / 4}{9.0}\right) \times 100 \\
& E_3=\left(\frac{0.1}{20.0}+\frac{0.2}{36.0}\right) \times 100=(0.005+0.00555 . . .) \times 100 \approx 1.055 \%
\end{aligned}
\)
Comparing the percentage errors, \(E_1 \approx 0.3125 \%, E_2 \approx 0.46875 \%\), and \(E_3 \approx 1.055 \%\). The minimum percentage error is obtained by student 1. - Question 18 of 20
18. Question
The radius of a sphere is measured to be \((7.50 \pm 0.85) \mathrm{cm}\). Suppose the percentage error in its volume is \(x\) . The value of \(x\), to the nearest \(x\), is _____. [JEE Main 2021 (Online) 18th March Evening Shift]
CorrectIncorrectHint
(d) Step 1: Calculate the percentage error in the radius
The formula for percentage error in the radius is \(\left(\frac{\Delta r}{r}\right) \times 100\).
Given values are \(\boldsymbol{r}=7.50 \mathrm{~cm}\) and \(\boldsymbol{\Delta} \boldsymbol{r}=0.85 \mathrm{~cm}\).
The percentage error in the radius is:
\(
\text { Percentage Error in Radius }=\left(\frac{0.85}{7.50}\right) \times 100 \approx 11.33 \%
\)
Step 2: Calculate the percentage error in the volume
The volume of a sphere is given by the formula \(V=\frac{4}{3} \pi r^3\).
For a quantity \(y\) that depends on \(r\) as \(y \propto r^n\), the percentage error in \(y\) is \(n\) times the percentage error in \(r\). In this case, \(n=3\).
The percentage error in the volume ( \(x\) ) is three times the percentage error in the radius:
\(x=3 \times\) Percentage Error in Radius
\(
\begin{gathered}
x=3 \times\left(\frac{0.85}{7.50}\right) \times 100 \\
x=\left(\frac{2.55}{7.50}\right) \times 100 \\
x=0.34 \times 100 \\
x=34 \%
\end{gathered}
\) - Question 19 of 20
19. Question
The resistance \(\mathrm{R}=\frac{V}{I}\), where \(\mathrm{V}=(50 \pm 2) \mathrm{V}\) and \(\mathrm{I}=(20 \pm 0.2) \mathrm{A}\). The percentage error in \(R\) is ‘ \(x\) ‘%. The value of ‘ \(x\) ‘ to the nearest integer is ____. [JEE Main 2021 (Online) 16th March Morning Shift]
CorrectIncorrectHint
(d)
\(
\begin{aligned}
& R=\frac{V}{I} \\
& \frac{\Delta R}{R} \times 100=\frac{\Delta V}{V} \times 100+\frac{\Delta I}{I} \times 100 \\
& \% \text { error in } R=\frac{2}{50} \times 100+\frac{0.2}{20} \times 100 \\
& \% \text { error in } \mathrm{R}=4+1 \\
& \therefore \% \text { error in } \mathrm{R}=5 \%
\end{aligned}
\) - Question 20 of 20
20. Question
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is \(\left(\frac{x}{100}\right) \%\). If the relative errors in measuring the mass and the diameter are \(6.0 \%\) and \(1.5 \%\) respectively, the value of \(x\) is _____. [JEE Main 2020 (Online) 6th September Morning Slot]
CorrectIncorrectHint
(d) Step 1: Express density in terms of mass and diameter
The density \(\rho\) of a sphere is given by the formula:
\(
\rho=\frac{\text { mass }}{\text { volume }}=\frac{m}{V}
\)
The volume \(V\) of a sphere with diameter \(d\) is:
\(
V=\frac{4}{3} \pi r^3=\frac{4}{3} \pi\left(\frac{d}{2}\right)^3=\frac{\pi d^3}{6}
\)
Substituting the volume formula into the density formula gives:
\(
\rho=\frac{m}{\frac{\pi d^3}{6}}=\frac{6 m}{\pi d^3}
\)
Step 2: Determine the maximum relative error
The formula for the maximum relative error in a quantity \(Y=\frac{A^a B^b}{C^c}\) is given by:
\(
\frac{\Delta Y}{Y}=a \frac{\Delta A}{A}+b \frac{\Delta B}{B}+c \frac{\Delta C}{C}
\)
For density \(\rho=\frac{6 m}{\pi d^3}\), the maximum relative error is:
\(
\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+3 \frac{\Delta d}{d}
\)
Step 3: Calculate the maximum percentage error
The percentage errors are given as:
Percentage error in mass, \(\left(\frac{\Delta m}{m} \times 100\right) \%=6.0 \%\)
Percentage error in diameter, \(\left(\frac{\Delta d}{d} \times 100\right) \%=1.5 \%\)
The maximum percentage error in density is:
\(
\begin{aligned}
& \left(\frac{\Delta \rho}{\rho} \times 100\right) \%=\left(\frac{\Delta m}{m} \times 100\right) \%+3 \times\left(\frac{\Delta d}{d} \times 100\right) \% \\
& \left(\frac{\Delta \rho}{\rho} \times 100\right) \%=6.0 \%+3 \times 1.5 \% \\
& \left(\frac{\Delta \rho}{\rho} \times 100\right) \%=6.0 \%+4.5 \%=10.5 \%
\end{aligned}
\)
The problem states that the maximum error in the density is \(\left(\frac{x}{100}\right) \%\). We calculated the maximum error to be \(10.5 \%\).
\(
\begin{gathered}
\left(\frac{x}{100}\right) \%=10.5 \% \\
\frac{x}{100}=10.5 \\
x=10.5 \times 100 \\
x=1050
\end{gathered}
\)
