Open App

JEE Exam > JEE Notes > Physics for JEE Main & Advanced > JEE Main Previous Year Questions (2025): Units & Measurements

JEE Main Previous Year Questions (2025): Units & Measurements | Physics for JEE Main & Advanced PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


JEE Main Previous Year Questions (2025): 
Units & Measurements 
Q1: The least count of a screw guage is 0.01 mm . If the pitch is increased by ???? % and number of 
divisions on the circular scale is reduced by ???? %, the new least count will be ____ × ????
-?? ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Answer: 35 
Solution: 
The least count (L.C.) of a screw gauge is given by: 
L.C. =
 Pitch 
 Number of Divisions on Circular Scale 
 
Let the original pitch be ?? and the original number of divisions be ?? . Then, the original least count is: 
?? ?? = 0.01 mm 
After modifications: 
The pitch is increased by 75%, so the new pitch is: 
?? new 
= ?? + 0.75?? = 1.75?? 
The number of divisions is reduced by 50%, so the new number of divisions is: 
?? new 
= 0.5?? 
The new least count is then: 
L.C
.new 
=
?? new 
?? new 
=
1.75?? 0.5?? =
1.75
0.5
×
?? ?? = 3.5 ×
?? ?? 
Substitute the original least count: 
L.C.  
new 
= 3.5× 0.01 mm= 0.035 mm 
Expressed in the form × 10
-3
 mm , this becomes: 
0.035 mm= 35× 10
-3
 mm 
Thus, the new least count is: 
35× 10
-3
 mm 
Q2: A tiny metallic rectangular sheet has length and breadth of ?? ???? and ?? .?? ???? , respectively. 
Using a specially designed screw gauge which has pitch of ?? .???? ???? and ???? divisions in the 
circular scale, you are asked to find the area of the sheet. In this measurement, the maximum 
fractional error will be 
?? ??????
 where ?? is ____ . 
JEE Main 2025 (Online) 28th January Morning Shift 
Answer: 3 
Solution: 
 
Page 2


JEE Main Previous Year Questions (2025): 
Units & Measurements 
Q1: The least count of a screw guage is 0.01 mm . If the pitch is increased by ???? % and number of 
divisions on the circular scale is reduced by ???? %, the new least count will be ____ × ????
-?? ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Answer: 35 
Solution: 
The least count (L.C.) of a screw gauge is given by: 
L.C. =
 Pitch 
 Number of Divisions on Circular Scale 
 
Let the original pitch be ?? and the original number of divisions be ?? . Then, the original least count is: 
?? ?? = 0.01 mm 
After modifications: 
The pitch is increased by 75%, so the new pitch is: 
?? new 
= ?? + 0.75?? = 1.75?? 
The number of divisions is reduced by 50%, so the new number of divisions is: 
?? new 
= 0.5?? 
The new least count is then: 
L.C
.new 
=
?? new 
?? new 
=
1.75?? 0.5?? =
1.75
0.5
×
?? ?? = 3.5 ×
?? ?? 
Substitute the original least count: 
L.C.  
new 
= 3.5× 0.01 mm= 0.035 mm 
Expressed in the form × 10
-3
 mm , this becomes: 
0.035 mm= 35× 10
-3
 mm 
Thus, the new least count is: 
35× 10
-3
 mm 
Q2: A tiny metallic rectangular sheet has length and breadth of ?? ???? and ?? .?? ???? , respectively. 
Using a specially designed screw gauge which has pitch of ?? .???? ???? and ???? divisions in the 
circular scale, you are asked to find the area of the sheet. In this measurement, the maximum 
fractional error will be 
?? ??????
 where ?? is ____ . 
JEE Main 2025 (Online) 28th January Morning Shift 
Answer: 3 
Solution: 
 
 
Given, L = 5 mm, B = 2.5 mm 
We know, least count of a screw gauge, 
L.C. =
 Pitch length 
 No. of division on circular scale 
 
? ?? .?? .=
0.75
15
= 0.05 mm 
We know, ?? = ???? 
By taking ???? on both sides, 
? ln ?? = ln ?? + ln ?? 
by differentiating both sides, 
?
????
?? =
????
?? +
????
?? 
Here, ???? = ???? = 0.05???? (L.C.) 
So fractional error, 
????
?? =
0.05
5
+
0.05
2.5
 
=
1
100
+
2
100
=
3
100
=
?? 100
 (given) 
Hence, ?? = 3. 
Q3: A physical quantity ?? is related to four observables ?? ,?? ,?? ,?? as follows : 
?? =
?? ?? ?? ????
 
where, 
?? = (???? ± ?? )???? ;?? = (???? ± ?? .?? )?? ;?? = (???? ± ?? .?? )??????
-?? and 
?? = (?? ?? ± ?? .?? )?? , then the percentage error in ?? is 
?? ????????
, where 
?? = ____ - 
JEE Main 2025 (Online) 29th January Evening Shift 
Answer: 7700 
Solution: 
Given, ?? =
?? ?? 4
????
 
?? = (60± 3)???? ? ?? = 60???? ,??? = 3???? 
?? = (20± 0.1)?? ? ?? = 20?? ,??? = 0.1?? 
?? = (40± 0.2)???? ?? -2
? ?? = 40???? ?? -2
,??? = 0.2???? ?? -2
 
?? = (50± 0.1)?? ? ?? = 50?? ,??? = 0.1?? 
As, ?? =
?? ?? 4
????
 
Page 3


JEE Main Previous Year Questions (2025): 
Units & Measurements 
Q1: The least count of a screw guage is 0.01 mm . If the pitch is increased by ???? % and number of 
divisions on the circular scale is reduced by ???? %, the new least count will be ____ × ????
-?? ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Answer: 35 
Solution: 
The least count (L.C.) of a screw gauge is given by: 
L.C. =
 Pitch 
 Number of Divisions on Circular Scale 
 
Let the original pitch be ?? and the original number of divisions be ?? . Then, the original least count is: 
?? ?? = 0.01 mm 
After modifications: 
The pitch is increased by 75%, so the new pitch is: 
?? new 
= ?? + 0.75?? = 1.75?? 
The number of divisions is reduced by 50%, so the new number of divisions is: 
?? new 
= 0.5?? 
The new least count is then: 
L.C
.new 
=
?? new 
?? new 
=
1.75?? 0.5?? =
1.75
0.5
×
?? ?? = 3.5 ×
?? ?? 
Substitute the original least count: 
L.C.  
new 
= 3.5× 0.01 mm= 0.035 mm 
Expressed in the form × 10
-3
 mm , this becomes: 
0.035 mm= 35× 10
-3
 mm 
Thus, the new least count is: 
35× 10
-3
 mm 
Q2: A tiny metallic rectangular sheet has length and breadth of ?? ???? and ?? .?? ???? , respectively. 
Using a specially designed screw gauge which has pitch of ?? .???? ???? and ???? divisions in the 
circular scale, you are asked to find the area of the sheet. In this measurement, the maximum 
fractional error will be 
?? ??????
 where ?? is ____ . 
JEE Main 2025 (Online) 28th January Morning Shift 
Answer: 3 
Solution: 
 
 
Given, L = 5 mm, B = 2.5 mm 
We know, least count of a screw gauge, 
L.C. =
 Pitch length 
 No. of division on circular scale 
 
? ?? .?? .=
0.75
15
= 0.05 mm 
We know, ?? = ???? 
By taking ???? on both sides, 
? ln ?? = ln ?? + ln ?? 
by differentiating both sides, 
?
????
?? =
????
?? +
????
?? 
Here, ???? = ???? = 0.05???? (L.C.) 
So fractional error, 
????
?? =
0.05
5
+
0.05
2.5
 
=
1
100
+
2
100
=
3
100
=
?? 100
 (given) 
Hence, ?? = 3. 
Q3: A physical quantity ?? is related to four observables ?? ,?? ,?? ,?? as follows : 
?? =
?? ?? ?? ????
 
where, 
?? = (???? ± ?? )???? ;?? = (???? ± ?? .?? )?? ;?? = (???? ± ?? .?? )??????
-?? and 
?? = (?? ?? ± ?? .?? )?? , then the percentage error in ?? is 
?? ????????
, where 
?? = ____ - 
JEE Main 2025 (Online) 29th January Evening Shift 
Answer: 7700 
Solution: 
Given, ?? =
?? ?? 4
????
 
?? = (60± 3)???? ? ?? = 60???? ,??? = 3???? 
?? = (20± 0.1)?? ? ?? = 20?? ,??? = 0.1?? 
?? = (40± 0.2)???? ?? -2
? ?? = 40???? ?? -2
,??? = 0.2???? ?? -2
 
?? = (50± 0.1)?? ? ?? = 50?? ,??? = 0.1?? 
As, ?? =
?? ?? 4
????
 
by taking ln on both sides, 
ln ?? = ln ?? + ?? ln ?? - ln ?? - ln ?? 
Now, by differentiating, 
????
?? =
????
?? + 4
????
?? -
????
?? -
????
?? 
So, maximum fractional error in Q is given by, 
??? ?? =
??? ?? + 4
??? ?? +
??? ?? +
??? ?? 
?
??? ?? =
3
60
+ 4(
0.1
20
) +
0.2
40
+
0.1
40
 
=
1
20
+
1
50
+
1
200
+
1
500
 
?
??? ?? =
50+ 20+ 5 + 2
1000
=
77
1000
 
Hence the % error in ?? =
??? ?? × 100% 
=
7700
1000
% =
?? 1000
 (given) 
So, ?? = 7700 
Q4: A physical quantity ?? is related to four other quantities ?? ,?? ,?? and s as follows 
?? =
?? ?? ?? ?? ?? v?? 
The percentage errors in the measurement of ?? ,?? ,?? and ?? are ?? %,?? %,?? % and ?? %, respectively. 
The percentage error in the measurement of ?? will be ____ % 
Answer: 15 
Solution: 
To determine the percentage error in the measurement of ?? , which is related to ?? ,?? ,?? , and ?? as: 
?? =
?? ?? 2
?? 3
v?? 
we first express it in terms of powers: 
?? = ?? 1
?? 2
?? -3
?? -1/2
 
The percentage error in ?? can be calculated using the formula for the propagation of error, which is: 
(
??? ?? )
max
=
??? ?? | + 2|
??? ?? | + 3|
??? ?? |+
1
2
|
??? ?? |  
Given the percentage errors for ?? ,?? ,?? , and ?? are 1%,2%,3%, and 2% respectively, we substitute these values into the 
formula: 
(
??? ?? )
max
= 1%+ 2 × 2% + 3 × 3% +
1
2
× 2% 
Calculating each part: 
Contribution from ?? :1% 
Contribution from ?? :4% (since 2 × 2% = 4% ) 
Contribution from ?? :9% (since 3 × 3% = 9% ) 
Contribution from ?? :1%( since 
1
2
× 2% = 1%) 
Adding these contributions together: 
1% + 4%+ 9% + 1% = 15% 
Thus, the maximum percentage error in the measurement of ?? is 15%. 
Page 4


JEE Main Previous Year Questions (2025): 
Units & Measurements 
Q1: The least count of a screw guage is 0.01 mm . If the pitch is increased by ???? % and number of 
divisions on the circular scale is reduced by ???? %, the new least count will be ____ × ????
-?? ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Answer: 35 
Solution: 
The least count (L.C.) of a screw gauge is given by: 
L.C. =
 Pitch 
 Number of Divisions on Circular Scale 
 
Let the original pitch be ?? and the original number of divisions be ?? . Then, the original least count is: 
?? ?? = 0.01 mm 
After modifications: 
The pitch is increased by 75%, so the new pitch is: 
?? new 
= ?? + 0.75?? = 1.75?? 
The number of divisions is reduced by 50%, so the new number of divisions is: 
?? new 
= 0.5?? 
The new least count is then: 
L.C
.new 
=
?? new 
?? new 
=
1.75?? 0.5?? =
1.75
0.5
×
?? ?? = 3.5 ×
?? ?? 
Substitute the original least count: 
L.C.  
new 
= 3.5× 0.01 mm= 0.035 mm 
Expressed in the form × 10
-3
 mm , this becomes: 
0.035 mm= 35× 10
-3
 mm 
Thus, the new least count is: 
35× 10
-3
 mm 
Q2: A tiny metallic rectangular sheet has length and breadth of ?? ???? and ?? .?? ???? , respectively. 
Using a specially designed screw gauge which has pitch of ?? .???? ???? and ???? divisions in the 
circular scale, you are asked to find the area of the sheet. In this measurement, the maximum 
fractional error will be 
?? ??????
 where ?? is ____ . 
JEE Main 2025 (Online) 28th January Morning Shift 
Answer: 3 
Solution: 
 
 
Given, L = 5 mm, B = 2.5 mm 
We know, least count of a screw gauge, 
L.C. =
 Pitch length 
 No. of division on circular scale 
 
? ?? .?? .=
0.75
15
= 0.05 mm 
We know, ?? = ???? 
By taking ???? on both sides, 
? ln ?? = ln ?? + ln ?? 
by differentiating both sides, 
?
????
?? =
????
?? +
????
?? 
Here, ???? = ???? = 0.05???? (L.C.) 
So fractional error, 
????
?? =
0.05
5
+
0.05
2.5
 
=
1
100
+
2
100
=
3
100
=
?? 100
 (given) 
Hence, ?? = 3. 
Q3: A physical quantity ?? is related to four observables ?? ,?? ,?? ,?? as follows : 
?? =
?? ?? ?? ????
 
where, 
?? = (???? ± ?? )???? ;?? = (???? ± ?? .?? )?? ;?? = (???? ± ?? .?? )??????
-?? and 
?? = (?? ?? ± ?? .?? )?? , then the percentage error in ?? is 
?? ????????
, where 
?? = ____ - 
JEE Main 2025 (Online) 29th January Evening Shift 
Answer: 7700 
Solution: 
Given, ?? =
?? ?? 4
????
 
?? = (60± 3)???? ? ?? = 60???? ,??? = 3???? 
?? = (20± 0.1)?? ? ?? = 20?? ,??? = 0.1?? 
?? = (40± 0.2)???? ?? -2
? ?? = 40???? ?? -2
,??? = 0.2???? ?? -2
 
?? = (50± 0.1)?? ? ?? = 50?? ,??? = 0.1?? 
As, ?? =
?? ?? 4
????
 
by taking ln on both sides, 
ln ?? = ln ?? + ?? ln ?? - ln ?? - ln ?? 
Now, by differentiating, 
????
?? =
????
?? + 4
????
?? -
????
?? -
????
?? 
So, maximum fractional error in Q is given by, 
??? ?? =
??? ?? + 4
??? ?? +
??? ?? +
??? ?? 
?
??? ?? =
3
60
+ 4(
0.1
20
) +
0.2
40
+
0.1
40
 
=
1
20
+
1
50
+
1
200
+
1
500
 
?
??? ?? =
50+ 20+ 5 + 2
1000
=
77
1000
 
Hence the % error in ?? =
??? ?? × 100% 
=
7700
1000
% =
?? 1000
 (given) 
So, ?? = 7700 
Q4: A physical quantity ?? is related to four other quantities ?? ,?? ,?? and s as follows 
?? =
?? ?? ?? ?? ?? v?? 
The percentage errors in the measurement of ?? ,?? ,?? and ?? are ?? %,?? %,?? % and ?? %, respectively. 
The percentage error in the measurement of ?? will be ____ % 
Answer: 15 
Solution: 
To determine the percentage error in the measurement of ?? , which is related to ?? ,?? ,?? , and ?? as: 
?? =
?? ?? 2
?? 3
v?? 
we first express it in terms of powers: 
?? = ?? 1
?? 2
?? -3
?? -1/2
 
The percentage error in ?? can be calculated using the formula for the propagation of error, which is: 
(
??? ?? )
max
=
??? ?? | + 2|
??? ?? | + 3|
??? ?? |+
1
2
|
??? ?? |  
Given the percentage errors for ?? ,?? ,?? , and ?? are 1%,2%,3%, and 2% respectively, we substitute these values into the 
formula: 
(
??? ?? )
max
= 1%+ 2 × 2% + 3 × 3% +
1
2
× 2% 
Calculating each part: 
Contribution from ?? :1% 
Contribution from ?? :4% (since 2 × 2% = 4% ) 
Contribution from ?? :9% (since 3 × 3% = 9% ) 
Contribution from ?? :1%( since 
1
2
× 2% = 1%) 
Adding these contributions together: 
1% + 4%+ 9% + 1% = 15% 
Thus, the maximum percentage error in the measurement of ?? is 15%. 
Q5: Given below are two statements : 
Statement I: In a vernier callipers, one vernier scale division is always smaller than one main 
scale division. 
Statement II : The vernier constant is given by one main scale division multiplied by the number 
of vernier scale divisions. 
In the light of the above statements, choose the correct answer from the options given below. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Statement I and Statement II are false 
B. Statement I is true but Statement II is false 
C. Both Statement I and Statement II are true 
D. Statement I is false but Statement II is true 
Answer: B 
Solution: 
In general, one vernier scale division is smaller than one main scale division but in some modified cases it may be not 
correct. Also least count is given by one main scale division divided by number of vernier scale division for normal 
vernier calliper. 
Hence, option 2 is correct. 
Q6: If ?? is magnetic field and ?? ?? is permeability of free space, then the dimensions of (?? /?? ?? ) is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. MT
-2
 A
-1
 
B. L
-1
 A 
C. ML
2
 T
-2
 A
-1
 
D. LT
-2
 A
-1
 
Answer: B 
Solution: 
To determine the dimensions of (
?? ?? 0
), we start with the formula for the magnetic field ?? inside a solenoid: 
?? = ?? 0
???? 
where ?? =
?? ?? is the number of turns per unit length, ?? is the current, and ?? 0
 is the permeability of free space. 
Rearranging the formula gives: 
?? ?? 0
= ???? =
?? ?? ?? 
Page 5


JEE Main Previous Year Questions (2025): 
Units & Measurements 
Q1: The least count of a screw guage is 0.01 mm . If the pitch is increased by ???? % and number of 
divisions on the circular scale is reduced by ???? %, the new least count will be ____ × ????
-?? ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Answer: 35 
Solution: 
The least count (L.C.) of a screw gauge is given by: 
L.C. =
 Pitch 
 Number of Divisions on Circular Scale 
 
Let the original pitch be ?? and the original number of divisions be ?? . Then, the original least count is: 
?? ?? = 0.01 mm 
After modifications: 
The pitch is increased by 75%, so the new pitch is: 
?? new 
= ?? + 0.75?? = 1.75?? 
The number of divisions is reduced by 50%, so the new number of divisions is: 
?? new 
= 0.5?? 
The new least count is then: 
L.C
.new 
=
?? new 
?? new 
=
1.75?? 0.5?? =
1.75
0.5
×
?? ?? = 3.5 ×
?? ?? 
Substitute the original least count: 
L.C.  
new 
= 3.5× 0.01 mm= 0.035 mm 
Expressed in the form × 10
-3
 mm , this becomes: 
0.035 mm= 35× 10
-3
 mm 
Thus, the new least count is: 
35× 10
-3
 mm 
Q2: A tiny metallic rectangular sheet has length and breadth of ?? ???? and ?? .?? ???? , respectively. 
Using a specially designed screw gauge which has pitch of ?? .???? ???? and ???? divisions in the 
circular scale, you are asked to find the area of the sheet. In this measurement, the maximum 
fractional error will be 
?? ??????
 where ?? is ____ . 
JEE Main 2025 (Online) 28th January Morning Shift 
Answer: 3 
Solution: 
 
 
Given, L = 5 mm, B = 2.5 mm 
We know, least count of a screw gauge, 
L.C. =
 Pitch length 
 No. of division on circular scale 
 
? ?? .?? .=
0.75
15
= 0.05 mm 
We know, ?? = ???? 
By taking ???? on both sides, 
? ln ?? = ln ?? + ln ?? 
by differentiating both sides, 
?
????
?? =
????
?? +
????
?? 
Here, ???? = ???? = 0.05???? (L.C.) 
So fractional error, 
????
?? =
0.05
5
+
0.05
2.5
 
=
1
100
+
2
100
=
3
100
=
?? 100
 (given) 
Hence, ?? = 3. 
Q3: A physical quantity ?? is related to four observables ?? ,?? ,?? ,?? as follows : 
?? =
?? ?? ?? ????
 
where, 
?? = (???? ± ?? )???? ;?? = (???? ± ?? .?? )?? ;?? = (???? ± ?? .?? )??????
-?? and 
?? = (?? ?? ± ?? .?? )?? , then the percentage error in ?? is 
?? ????????
, where 
?? = ____ - 
JEE Main 2025 (Online) 29th January Evening Shift 
Answer: 7700 
Solution: 
Given, ?? =
?? ?? 4
????
 
?? = (60± 3)???? ? ?? = 60???? ,??? = 3???? 
?? = (20± 0.1)?? ? ?? = 20?? ,??? = 0.1?? 
?? = (40± 0.2)???? ?? -2
? ?? = 40???? ?? -2
,??? = 0.2???? ?? -2
 
?? = (50± 0.1)?? ? ?? = 50?? ,??? = 0.1?? 
As, ?? =
?? ?? 4
????
 
by taking ln on both sides, 
ln ?? = ln ?? + ?? ln ?? - ln ?? - ln ?? 
Now, by differentiating, 
????
?? =
????
?? + 4
????
?? -
????
?? -
????
?? 
So, maximum fractional error in Q is given by, 
??? ?? =
??? ?? + 4
??? ?? +
??? ?? +
??? ?? 
?
??? ?? =
3
60
+ 4(
0.1
20
) +
0.2
40
+
0.1
40
 
=
1
20
+
1
50
+
1
200
+
1
500
 
?
??? ?? =
50+ 20+ 5 + 2
1000
=
77
1000
 
Hence the % error in ?? =
??? ?? × 100% 
=
7700
1000
% =
?? 1000
 (given) 
So, ?? = 7700 
Q4: A physical quantity ?? is related to four other quantities ?? ,?? ,?? and s as follows 
?? =
?? ?? ?? ?? ?? v?? 
The percentage errors in the measurement of ?? ,?? ,?? and ?? are ?? %,?? %,?? % and ?? %, respectively. 
The percentage error in the measurement of ?? will be ____ % 
Answer: 15 
Solution: 
To determine the percentage error in the measurement of ?? , which is related to ?? ,?? ,?? , and ?? as: 
?? =
?? ?? 2
?? 3
v?? 
we first express it in terms of powers: 
?? = ?? 1
?? 2
?? -3
?? -1/2
 
The percentage error in ?? can be calculated using the formula for the propagation of error, which is: 
(
??? ?? )
max
=
??? ?? | + 2|
??? ?? | + 3|
??? ?? |+
1
2
|
??? ?? |  
Given the percentage errors for ?? ,?? ,?? , and ?? are 1%,2%,3%, and 2% respectively, we substitute these values into the 
formula: 
(
??? ?? )
max
= 1%+ 2 × 2% + 3 × 3% +
1
2
× 2% 
Calculating each part: 
Contribution from ?? :1% 
Contribution from ?? :4% (since 2 × 2% = 4% ) 
Contribution from ?? :9% (since 3 × 3% = 9% ) 
Contribution from ?? :1%( since 
1
2
× 2% = 1%) 
Adding these contributions together: 
1% + 4%+ 9% + 1% = 15% 
Thus, the maximum percentage error in the measurement of ?? is 15%. 
Q5: Given below are two statements : 
Statement I: In a vernier callipers, one vernier scale division is always smaller than one main 
scale division. 
Statement II : The vernier constant is given by one main scale division multiplied by the number 
of vernier scale divisions. 
In the light of the above statements, choose the correct answer from the options given below. 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Statement I and Statement II are false 
B. Statement I is true but Statement II is false 
C. Both Statement I and Statement II are true 
D. Statement I is false but Statement II is true 
Answer: B 
Solution: 
In general, one vernier scale division is smaller than one main scale division but in some modified cases it may be not 
correct. Also least count is given by one main scale division divided by number of vernier scale division for normal 
vernier calliper. 
Hence, option 2 is correct. 
Q6: If ?? is magnetic field and ?? ?? is permeability of free space, then the dimensions of (?? /?? ?? ) is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. MT
-2
 A
-1
 
B. L
-1
 A 
C. ML
2
 T
-2
 A
-1
 
D. LT
-2
 A
-1
 
Answer: B 
Solution: 
To determine the dimensions of (
?? ?? 0
), we start with the formula for the magnetic field ?? inside a solenoid: 
?? = ?? 0
???? 
where ?? =
?? ?? is the number of turns per unit length, ?? is the current, and ?? 0
 is the permeability of free space. 
Rearranging the formula gives: 
?? ?? 0
= ???? =
?? ?? ?? 
From the above, the dimensions of (
?? ?? 0
) are: 
[
?? ?? 0
] = [?? -1
?? ] 
Thus, the dimensional formula of (
?? ?? 0
) corresponds to ?? -1
 A. 
Q7: The maximum percentage error in the measurment of density of a wire is 
[Given, mass of wire = (?? .???? ± ?? .?????? )?? 
radius of wire = (?? .???? ± ?? .???? )???? 
length of wire = (???? .???? ± ?? .???? )???? ] 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 7 
B. 8 
C. 5 
D. 4 
Answer: C 
Solution: 
To determine the maximum percentage error in the density of the wire, follow these steps: 
The density of the wire is given by the formula for a cylinder: 
?? =
?? ?? ?? 2
?? 
When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into 
account the power to which each variable is raised. Thus, for density: 
??? ?? =
??? ?? + 2
??? ?? +
??? ?? 
Here: 
??? ?? is the relative error in the mass. 
2
??? ?? is due to the radius being squared. 
??? ?? is the relative error in the length. 
Now plug in the given values: 
Mass: ?? = 0.60 g with an error ??? = 0.003 g 
??? ?? =
0.003
0.60
= 0.005 or 0.5%. 
Radius: ?? = 0.50 cm with an error ??? = 0.01 cm 
??? ?? =
0.01
0.50
= 0.02 or 2%. Since the radius is squared in the formula, multiply by 2 : 
2
??? ?? = 2 × 0.02 = 0.04 or 4%. 
Length: ?? = 10.00 cm with an error ??? = 0.05 cm 
??? ?? =
0.05
10.00
= 0.005 or 0.5%. 
Sum these contributions to find the overall maximum relative error: 
??? ?? = 0.005+ 0.04+ 0.005= 0.05

	Physics for JEE Main & Advanced 268 videos\|740 docs\|171 tests

Physics for JEE Main & Advanced

268 videos|740 docs|171 tests

Join Course for Free

FAQs on JEE Main Previous Year Questions (2025): Units & Measurements - Physics for JEE Main & Advanced

1. What are the fundamental units of measurement in the International System of Units (SI) ?

Ans. The International System of Units (SI) defines seven fundamental units of measurement. These are: 1. Meter (m) for length 2. Kilogram (kg) for mass 3. Second (s) for time 4. Ampere (A) for electric current 5. Kelvin (K) for temperature 6. Mole (mol) for the amount of substance 7. Candela (cd) for luminous intensity. These units serve as the foundation for all other derived units used in scientific measurements.

2. How do you convert units from one system to another, such as from metric to imperial ?

Ans. To convert units from one system to another, you need to know the conversion factors between the two systems. For instance, to convert meters to feet, you multiply the length in meters by approximately 3.28084 (since 1 meter equals about 3.28084 feet). Similarly, for mass, to convert kilograms to pounds, you multiply by approximately 2.20462. Always ensure to keep track of units and apply the appropriate conversion factor for accurate results.

3. What is the significance of significant figures in measurements ?

Ans. Significant figures are important in measurements as they convey the precision of the measured value. They include all certain digits plus one estimated digit. The rules for determining significant figures help in maintaining consistency and accuracy in calculations and reporting results. For example, if a measurement is reported as 12.30 cm, it indicates that the measurement is precise to the hundredths place, reflecting more accuracy than a measurement given as 12 cm.

4. What is the difference between accuracy and precision in measurements ?

Ans. Accuracy refers to how close a measured value is to the true or accepted value, while precision indicates the reproducibility of measurements, or how closely multiple measurements agree with each other. For example, if a set of measurements is consistently close to a known value, it is accurate. If multiple measurements yield the same result repeatedly, they are precise. A measurement can be precise but not accurate if it consistently misses the true value.

5. How can errors in measurements be minimized in experiments ?

Ans. Errors in measurements can be minimized through various techniques: 1. Calibration of instruments before use to ensure they provide accurate readings. 2. Repeating measurements multiple times and averaging the results to reduce random errors. 3. Using appropriate techniques and procedures to minimize systematic errors, such as ensuring proper alignment and environmental controls. 4. Being aware of and correcting for human errors during the measurement process. Implementing these strategies can greatly enhance the reliability of experimental results.

About this Document

Oct 01, 2025 Last updated

Related Exams

JEE

Document Description: JEE Main Previous Year Questions (2025): Units & Measurements for JEE 2025 is part of Physics for JEE Main & Advanced preparation. The notes and questions for JEE Main Previous Year Questions (2025): Units & Measurements have been prepared according to the JEE exam syllabus. Information about JEE Main Previous Year Questions (2025): Units & Measurements covers topics like and JEE Main Previous Year Questions (2025): Units & Measurements Example, for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main Previous Year Questions (2025): Units & Measurements.

Introduction of JEE Main Previous Year Questions (2025): Units & Measurements in English is available as part of our Physics for JEE Main & Advanced for JEE & JEE Main Previous Year Questions (2025): Units & Measurements in Hindi for Physics for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Main Previous Year Questions (2025): Units & Measurements | Physics for JEE Main & Advanced

Description

Full syllabus notes, lecture & questions for JEE Main Previous Year Questions (2025): Units & Measurements | Physics for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Physics for JEE Main & Advanced | Best notes, free PDF download

Information about JEE Main Previous Year Questions (2025): Units & Measurements

In this doc you can find the meaning of JEE Main Previous Year Questions (2025): Units & Measurements defined & explained in the simplest way possible. Besides explaining types of JEE Main Previous Year Questions (2025): Units & Measurements theory, EduRev gives you an ample number of questions to practice JEE Main Previous Year Questions (2025): Units & Measurements tests, examples and also practice JEE tests

	Physics for JEE Main & Advanced 268 videos\|740 docs\|171 tests

Physics for JEE Main & Advanced

268 videos|740 docs|171 tests

Join Course for Free

Download as PDF

Explore Courses for JEE exam

study material

ppt

mock tests for examination

Sample Paper

pdf

shortcuts and tricks

past year papers

Viva Questions

Free

video lectures

Exam

MCQs

Previous Year Questions with Solutions

practice quizzes

Semester Notes

Objective type Questions

Summary

JEE Main Previous Year Questions (2025): Units & Measurements | Physics for JEE Main & Advanced

Extra Questions

JEE Main Previous Year Questions (2025): Units & Measurements | Physics for JEE Main & Advanced

Important questions

;

Additional Information about JEE Main Previous Year Questions (2025): Units & Measurements for JEE Preparation

JEE Main Previous Year Questions (2025): Units & Measurements Free PDF Download

The JEE Main Previous Year Questions (2025): Units & Measurements is an invaluable resource that delves deep into the core of the JEE exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the JEE Main Previous Year Questions (2025): Units & Measurements now and kickstart your journey towards success in the JEE exam.

Importance of JEE Main Previous Year Questions (2025): Units & Measurements

The importance of JEE Main Previous Year Questions (2025): Units & Measurements cannot be overstated, especially for JEE aspirants. This document holds the key to success in the JEE exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

JEE Main Previous Year Questions (2025): Units & Measurements Notes

JEE Main Previous Year Questions (2025): Units & Measurements Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to JEE Main Previous Year Questions (2025): Units & Measurements. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, JEE Main Previous Year Questions (2025): Units & Measurements Notes on EduRev are your ultimate resource for success.

JEE Main Previous Year Questions (2025): Units & Measurements JEE

The "JEE Main Previous Year Questions (2025): Units & Measurements JEE Questions" guide is a valuable resource for all aspiring students preparing for the JEE exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study JEE Main Previous Year Questions (2025): Units & Measurements on the App

Students of JEE can study JEE Main Previous Year Questions (2025): Units & Measurements alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the JEE Main Previous Year Questions (2025): Units & Measurements, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of JEE Main Previous Year Questions (2025): Units & Measurements is prepared as per the latest JEE syllabus.

Education Revolution