JEE Mains Previous Year Questions (2021-2024): Units & Measurements | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Units & Measurements 
2024 
Q1: Match List - I with List - II. 
List I (Number) List II (Significant figure) 
(A) 1001 (I) 3 
(B) 010.1 (II) 4 
(C) 100.100 (III) 5 
(D) 0.0010010 (IV) 6 
Choose the correct answer from the options given below : 
(a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
(b) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
(c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 
(d) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Significant figures in a number represent the digits that carry meaning contributing to its precision. This 
includes all digits except: 
All leading zeros. 
Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must 
not be significant if there is no decimal point). 
Here is the explanation for the significant figures of each number in List I: 
(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting 
as placeholders. Hence, this number has 4 significant figures. 
(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 
1itself, andthe1 after the decimal point are all significant. So, this number has 3 significant figures. 
(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits 
or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 
significant figures. 
(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the 
number are significant because they are between significant digits or at the end of the number after the 
decimal. Therefore, this number has 5 significant figures. 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Units & Measurements 
2024 
Q1: Match List - I with List - II. 
List I (Number) List II (Significant figure) 
(A) 1001 (I) 3 
(B) 010.1 (II) 4 
(C) 100.100 (III) 5 
(D) 0.0010010 (IV) 6 
Choose the correct answer from the options given below : 
(a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
(b) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
(c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 
(d) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Significant figures in a number represent the digits that carry meaning contributing to its precision. This 
includes all digits except: 
All leading zeros. 
Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must 
not be significant if there is no decimal point). 
Here is the explanation for the significant figures of each number in List I: 
(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting 
as placeholders. Hence, this number has 4 significant figures. 
(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 
1itself, andthe1 after the decimal point are all significant. So, this number has 3 significant figures. 
(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits 
or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 
significant figures. 
(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the 
number are significant because they are between significant digits or at the end of the number after the 
decimal. Therefore, this number has 5 significant figures. 
Given List I and List II, the correct matching based on the explanation of significant figures is as follows: 
Option A 
(A) - II (1001 has 4 significant figures) 
(B) - I (010.1 has 3 significant figures) 
(C) - IV (100.100 has 6 significant figures) 
(D) - III ( 0.0010010 has 5 significant figures) 
Thus, the correct answer is Option A. 
Q2: 10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. 
If each division on the main scale is of 5 units, the least count of the instrument is : 
(a)  
?? ????
 
(b)  
????
????
 
(c)  
????
????
 
(d)  
?? ??      [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is 
determined by the difference in the measurement of one main scale division and one Vernier scale 
division. 
In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means 
that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each 
division on the main scale is 5 units, this is equal to: 
10 divisions on main scale × 5 units per division = 50 units 
The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions 
on the Vernier scale. This means that: 
1 division on the Vernier scale =
50 units 
11
 
Therefore, the least count of the Vernier caliper, which is the difference between one division on the 
main scale and one division on the Vernier scale, can be calculated as follows: 
Least count = value of one main scale division - value of one Vernier scale division 
Least count = 5 units -
50 units 
11
 
Least count =
55 units 
11
-
50 units 
11
 
Least count =
5 units 
11
 
Thus, the least count of the Vernier caliper is 
5 units 
11
 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Units & Measurements 
2024 
Q1: Match List - I with List - II. 
List I (Number) List II (Significant figure) 
(A) 1001 (I) 3 
(B) 010.1 (II) 4 
(C) 100.100 (III) 5 
(D) 0.0010010 (IV) 6 
Choose the correct answer from the options given below : 
(a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
(b) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
(c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 
(d) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Significant figures in a number represent the digits that carry meaning contributing to its precision. This 
includes all digits except: 
All leading zeros. 
Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must 
not be significant if there is no decimal point). 
Here is the explanation for the significant figures of each number in List I: 
(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting 
as placeholders. Hence, this number has 4 significant figures. 
(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 
1itself, andthe1 after the decimal point are all significant. So, this number has 3 significant figures. 
(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits 
or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 
significant figures. 
(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the 
number are significant because they are between significant digits or at the end of the number after the 
decimal. Therefore, this number has 5 significant figures. 
Given List I and List II, the correct matching based on the explanation of significant figures is as follows: 
Option A 
(A) - II (1001 has 4 significant figures) 
(B) - I (010.1 has 3 significant figures) 
(C) - IV (100.100 has 6 significant figures) 
(D) - III ( 0.0010010 has 5 significant figures) 
Thus, the correct answer is Option A. 
Q2: 10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. 
If each division on the main scale is of 5 units, the least count of the instrument is : 
(a)  
?? ????
 
(b)  
????
????
 
(c)  
????
????
 
(d)  
?? ??      [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is 
determined by the difference in the measurement of one main scale division and one Vernier scale 
division. 
In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means 
that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each 
division on the main scale is 5 units, this is equal to: 
10 divisions on main scale × 5 units per division = 50 units 
The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions 
on the Vernier scale. This means that: 
1 division on the Vernier scale =
50 units 
11
 
Therefore, the least count of the Vernier caliper, which is the difference between one division on the 
main scale and one division on the Vernier scale, can be calculated as follows: 
Least count = value of one main scale division - value of one Vernier scale division 
Least count = 5 units -
50 units 
11
 
Least count =
55 units 
11
-
50 units 
11
 
Least count =
5 units 
11
 
Thus, the least count of the Vernier caliper is 
5 units 
11
 
Therefore, the correct answer is: 
Option A: 
5
11
 
Q3: The dimensional formula of angular impulse is: 
(a) [????
?? ?? -?? ] 
(b) [????
-?? ?? -?? ] 
(c) [????
?? ?? -?? ] 
(d) [?????? -?? ]       [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Angular impulse is given when a torque is applied for a certain amount of time. The angular impulse 
changes the angular momentum of an object and has the same dimensions as angular momentum. 
The dimensional formula for torque ?? is the same as that for work, since torque is a kind of rotational 
work, which is given by force times distance (or in rotational terms, it can be considered as force times 
lever arm). The dimensional formula for force is [MLT
-2
], and when multiplied by distance [L], we get: 
[ Torque ] = [MLT
-2
] × [L] = [ML
2
 T
-2
] 
Now, angular impulse is torque times time, so we multiply the dimension of torque by time [T] : 
[ Angular Impulse ] = [ML
2
 T
-2
] × [T] = [ML
2
 T
-1
] 
So the correct dimensional formula for angular impulse is given by Option C, which is [ML
2
 T
-1
]. 
Q4: The radius ( ?? ), length (?? ) and resistance (R) of a metal wire was measured in the laboratory as 
 
?? = (?? . ???? ± ?? . ???? )???? 
?? = (?????? ± ???? )?????? 
?? = (???? ± ?? . ?? )???? 
The percentage error in resistivity of the material of the wire is : 
(a) ?? ?? . ?? % 
(b) ???? . ?? % 
(c) ???? . ?? % 
(d) ???? . ?? %        [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (d) 
To calculate the percentage error in the resistivity of the material of the wire, we need to understand 
the formula for resistivity. The resistivity ?? of a wire is given by: 
?? =
????
?? 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Units & Measurements 
2024 
Q1: Match List - I with List - II. 
List I (Number) List II (Significant figure) 
(A) 1001 (I) 3 
(B) 010.1 (II) 4 
(C) 100.100 (III) 5 
(D) 0.0010010 (IV) 6 
Choose the correct answer from the options given below : 
(a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
(b) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
(c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 
(d) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Significant figures in a number represent the digits that carry meaning contributing to its precision. This 
includes all digits except: 
All leading zeros. 
Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must 
not be significant if there is no decimal point). 
Here is the explanation for the significant figures of each number in List I: 
(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting 
as placeholders. Hence, this number has 4 significant figures. 
(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 
1itself, andthe1 after the decimal point are all significant. So, this number has 3 significant figures. 
(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits 
or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 
significant figures. 
(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the 
number are significant because they are between significant digits or at the end of the number after the 
decimal. Therefore, this number has 5 significant figures. 
Given List I and List II, the correct matching based on the explanation of significant figures is as follows: 
Option A 
(A) - II (1001 has 4 significant figures) 
(B) - I (010.1 has 3 significant figures) 
(C) - IV (100.100 has 6 significant figures) 
(D) - III ( 0.0010010 has 5 significant figures) 
Thus, the correct answer is Option A. 
Q2: 10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. 
If each division on the main scale is of 5 units, the least count of the instrument is : 
(a)  
?? ????
 
(b)  
????
????
 
(c)  
????
????
 
(d)  
?? ??      [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is 
determined by the difference in the measurement of one main scale division and one Vernier scale 
division. 
In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means 
that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each 
division on the main scale is 5 units, this is equal to: 
10 divisions on main scale × 5 units per division = 50 units 
The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions 
on the Vernier scale. This means that: 
1 division on the Vernier scale =
50 units 
11
 
Therefore, the least count of the Vernier caliper, which is the difference between one division on the 
main scale and one division on the Vernier scale, can be calculated as follows: 
Least count = value of one main scale division - value of one Vernier scale division 
Least count = 5 units -
50 units 
11
 
Least count =
55 units 
11
-
50 units 
11
 
Least count =
5 units 
11
 
Thus, the least count of the Vernier caliper is 
5 units 
11
 
Therefore, the correct answer is: 
Option A: 
5
11
 
Q3: The dimensional formula of angular impulse is: 
(a) [????
?? ?? -?? ] 
(b) [????
-?? ?? -?? ] 
(c) [????
?? ?? -?? ] 
(d) [?????? -?? ]       [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Angular impulse is given when a torque is applied for a certain amount of time. The angular impulse 
changes the angular momentum of an object and has the same dimensions as angular momentum. 
The dimensional formula for torque ?? is the same as that for work, since torque is a kind of rotational 
work, which is given by force times distance (or in rotational terms, it can be considered as force times 
lever arm). The dimensional formula for force is [MLT
-2
], and when multiplied by distance [L], we get: 
[ Torque ] = [MLT
-2
] × [L] = [ML
2
 T
-2
] 
Now, angular impulse is torque times time, so we multiply the dimension of torque by time [T] : 
[ Angular Impulse ] = [ML
2
 T
-2
] × [T] = [ML
2
 T
-1
] 
So the correct dimensional formula for angular impulse is given by Option C, which is [ML
2
 T
-1
]. 
Q4: The radius ( ?? ), length (?? ) and resistance (R) of a metal wire was measured in the laboratory as 
 
?? = (?? . ???? ± ?? . ???? )???? 
?? = (?????? ± ???? )?????? 
?? = (???? ± ?? . ?? )???? 
The percentage error in resistivity of the material of the wire is : 
(a) ?? ?? . ?? % 
(b) ???? . ?? % 
(c) ???? . ?? % 
(d) ???? . ?? %        [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (d) 
To calculate the percentage error in the resistivity of the material of the wire, we need to understand 
the formula for resistivity. The resistivity ?? of a wire is given by: 
?? =
????
?? 
where: 
?? is the resistance 
?? is the cross-sectional area of the wire 
?? is the length of the wire 
The cross-sectional area ?? of the wire with radius ?? is: 
?? = ?? ?? 2
 
We can plug this into the equation for resistivity to get: 
?? =
???? ?? 2
?? 
Now, to find the percentage error in resistivity, we need to find the percentage errors in ?? , ?? , and ?? and 
then use the following rule for combining errors: 
For a given function, ?? = ?? (?? , ?? , ?? , … ), where ?? , ?? , ?? , … are the measured quantities with possible errors, 
the percentage error in ?? , denoted as (???? )
%
, can be approximated by adding the relative percentage 
errors of the input quantities. If ?? has the form of a product and quotient of the measured quantities as 
in our case ( ?? =
???? ?? 2
?? ), the percentage error in ?? is given by: 
(???? )
%
= (???? )
%
+ (???? )
%
+ (???? )
%
+ ? 
Where (???? )
%
, (???? )
%
, and (???? )
%
 are the percentage errors in each measured quantity ?? , ?? , ?? , … 
respectively. 
For our case: 
The percentage error in radius (???? )
%
 is given by the error in ?? divided by the average radius and then 
multiplied by 100 : 
(???? )
%
= (
0.05
0.35
) × 100 
The percentage error in resistance (???? )
%
 is: 
(???? )
%
= (
10
100
) × 100 
The percentage error in resistance (???? )
%
 is: 
(???? )
%
= (
10
100
) × 100 
The percentage error in length (???? )
%
 is: 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Units & Measurements 
2024 
Q1: Match List - I with List - II. 
List I (Number) List II (Significant figure) 
(A) 1001 (I) 3 
(B) 010.1 (II) 4 
(C) 100.100 (III) 5 
(D) 0.0010010 (IV) 6 
Choose the correct answer from the options given below : 
(a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
(b) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
(c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 
(d) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Significant figures in a number represent the digits that carry meaning contributing to its precision. This 
includes all digits except: 
All leading zeros. 
Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must 
not be significant if there is no decimal point). 
Here is the explanation for the significant figures of each number in List I: 
(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting 
as placeholders. Hence, this number has 4 significant figures. 
(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 
1itself, andthe1 after the decimal point are all significant. So, this number has 3 significant figures. 
(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits 
or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 
significant figures. 
(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the 
number are significant because they are between significant digits or at the end of the number after the 
decimal. Therefore, this number has 5 significant figures. 
Given List I and List II, the correct matching based on the explanation of significant figures is as follows: 
Option A 
(A) - II (1001 has 4 significant figures) 
(B) - I (010.1 has 3 significant figures) 
(C) - IV (100.100 has 6 significant figures) 
(D) - III ( 0.0010010 has 5 significant figures) 
Thus, the correct answer is Option A. 
Q2: 10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. 
If each division on the main scale is of 5 units, the least count of the instrument is : 
(a)  
?? ????
 
(b)  
????
????
 
(c)  
????
????
 
(d)  
?? ??      [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is 
determined by the difference in the measurement of one main scale division and one Vernier scale 
division. 
In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means 
that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each 
division on the main scale is 5 units, this is equal to: 
10 divisions on main scale × 5 units per division = 50 units 
The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions 
on the Vernier scale. This means that: 
1 division on the Vernier scale =
50 units 
11
 
Therefore, the least count of the Vernier caliper, which is the difference between one division on the 
main scale and one division on the Vernier scale, can be calculated as follows: 
Least count = value of one main scale division - value of one Vernier scale division 
Least count = 5 units -
50 units 
11
 
Least count =
55 units 
11
-
50 units 
11
 
Least count =
5 units 
11
 
Thus, the least count of the Vernier caliper is 
5 units 
11
 
Therefore, the correct answer is: 
Option A: 
5
11
 
Q3: The dimensional formula of angular impulse is: 
(a) [????
?? ?? -?? ] 
(b) [????
-?? ?? -?? ] 
(c) [????
?? ?? -?? ] 
(d) [?????? -?? ]       [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Angular impulse is given when a torque is applied for a certain amount of time. The angular impulse 
changes the angular momentum of an object and has the same dimensions as angular momentum. 
The dimensional formula for torque ?? is the same as that for work, since torque is a kind of rotational 
work, which is given by force times distance (or in rotational terms, it can be considered as force times 
lever arm). The dimensional formula for force is [MLT
-2
], and when multiplied by distance [L], we get: 
[ Torque ] = [MLT
-2
] × [L] = [ML
2
 T
-2
] 
Now, angular impulse is torque times time, so we multiply the dimension of torque by time [T] : 
[ Angular Impulse ] = [ML
2
 T
-2
] × [T] = [ML
2
 T
-1
] 
So the correct dimensional formula for angular impulse is given by Option C, which is [ML
2
 T
-1
]. 
Q4: The radius ( ?? ), length (?? ) and resistance (R) of a metal wire was measured in the laboratory as 
 
?? = (?? . ???? ± ?? . ???? )???? 
?? = (?????? ± ???? )?????? 
?? = (???? ± ?? . ?? )???? 
The percentage error in resistivity of the material of the wire is : 
(a) ?? ?? . ?? % 
(b) ???? . ?? % 
(c) ???? . ?? % 
(d) ???? . ?? %        [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (d) 
To calculate the percentage error in the resistivity of the material of the wire, we need to understand 
the formula for resistivity. The resistivity ?? of a wire is given by: 
?? =
????
?? 
where: 
?? is the resistance 
?? is the cross-sectional area of the wire 
?? is the length of the wire 
The cross-sectional area ?? of the wire with radius ?? is: 
?? = ?? ?? 2
 
We can plug this into the equation for resistivity to get: 
?? =
???? ?? 2
?? 
Now, to find the percentage error in resistivity, we need to find the percentage errors in ?? , ?? , and ?? and 
then use the following rule for combining errors: 
For a given function, ?? = ?? (?? , ?? , ?? , … ), where ?? , ?? , ?? , … are the measured quantities with possible errors, 
the percentage error in ?? , denoted as (???? )
%
, can be approximated by adding the relative percentage 
errors of the input quantities. If ?? has the form of a product and quotient of the measured quantities as 
in our case ( ?? =
???? ?? 2
?? ), the percentage error in ?? is given by: 
(???? )
%
= (???? )
%
+ (???? )
%
+ (???? )
%
+ ? 
Where (???? )
%
, (???? )
%
, and (???? )
%
 are the percentage errors in each measured quantity ?? , ?? , ?? , … 
respectively. 
For our case: 
The percentage error in radius (???? )
%
 is given by the error in ?? divided by the average radius and then 
multiplied by 100 : 
(???? )
%
= (
0.05
0.35
) × 100 
The percentage error in resistance (???? )
%
 is: 
(???? )
%
= (
10
100
) × 100 
The percentage error in resistance (???? )
%
 is: 
(???? )
%
= (
10
100
) × 100 
The percentage error in length (???? )
%
 is: 
(???? )
%
= (
0.2
15
) × 100 
Now let's calculate each: 
 (???? )
%
= (
0.05
0.35
) × 100 ˜ 14.29%
 (???? )
%
= (
10
100
) × 100 = 10%
 (???? )
%
= (
0.2
15
) × 100 ˜ 1.33%
 
However, since the area ?? is proportional to ?? 2
, the percentage error in ?? will be twice the percentage 
error in ?? . Thus: 
(???? )
%
= 2 × (???? )
%
= 2 × 14.29% ˜ 28.58% 
Finally, we add the percentage errors to find the percentage error in resistivity: 
 (???? )
%
= (???? )
%
+ (???? )
%
+ (???? )
%
 (???? )
%
= 10% + 28.58% + 1.33% ˜ 39.91%
 
This calculation gives us a value close to 39.91%, which means the correct option is closest to this value. 
Thus, the best answer is: 
Option D 39.9% 
Q5: Consider two physical quantities ?? and ?? related to each other as ?? =
?? -?? ?? ????
 where ?? , ?? and ?? have 
dimensions of energy, length and time respectively. The dimension of ???? is 
(a) ?? -?? ?? ?? ?? ?? 
(b) ?? ?? ?? -?? ?? ?? 
(c) ?? ?? ?? -?? ?? ?? 
(d) ?? -?? ?? -?? ?? ??        [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
[B] = L
2
 A =
x
2
tE
=
L
2
TML
2
 T
-2
=
1
MT
-1
[ A] = M
-1 T
[AB] = [L
2
M
-1
 T
1
]
 
Q6: The measured value of the length of a simple pendulum is ???? ???? with ?? ???? accuracy. The time 
for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these 
measurements, the accuracy in the measurement of acceleration due to gravity is ?? %. The value of ?? 
is: 
(a) 6 
(b) 5 
(c) 4