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1. (a) A and B have different dimensions. Hence, only A and B in a ratio
form and is meaningful.
2. (d) No of divisions on main scale = N
No of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) ? b = 
Least count is a – b = a – 
=  = 
3. (a)
4. (a) T = P
a
 D
b
 S
c
M
0
L
0
T
1
 = (ML
–1
 T
–2
)
a
 (ML
–3
)
b
 (MT
–2
)
c
= M
a+b+c
 L
–a–3b
 T
–2a–2c
Applying principle of homogeneity
a + b + c = 0;  – a – 3b = 0;  – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
5. (b) As 
 
6. (a) Here, Mass of a body, M = 5.00 ± 0.05 kg
Volume of a body, V = 1.00 ± 0.05 m
3
Density, ? = 
Page 2


1. (a) A and B have different dimensions. Hence, only A and B in a ratio
form and is meaningful.
2. (d) No of divisions on main scale = N
No of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) ? b = 
Least count is a – b = a – 
=  = 
3. (a)
4. (a) T = P
a
 D
b
 S
c
M
0
L
0
T
1
 = (ML
–1
 T
–2
)
a
 (ML
–3
)
b
 (MT
–2
)
c
= M
a+b+c
 L
–a–3b
 T
–2a–2c
Applying principle of homogeneity
a + b + c = 0;  – a – 3b = 0;  – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
5. (b) As 
 
6. (a) Here, Mass of a body, M = 5.00 ± 0.05 kg
Volume of a body, V = 1.00 ± 0.05 m
3
Density, ? = 
Relative error in density is,  
Percentage error in density is
 = 1% + 5% = 6%
7. (d) For angular momentum, the dimensional formula is [ML
2
T
–1
]. For
other three, it is [ML
2
T
–2
].
8. (d) Let X = [ML
–1
 T
–1
]
Then,  
As we know,  
= (1 + 1.5 + 3) % = 5.5 %.
9. (d) Let unit ‘u’ related with e, a
0
, h and c as follows.
[u] = [e]
a
 [a
0
]
b
 [h]
c 
[C]
d
Using dimensional method,
[M
–1
L
–2
T
+4
A
+2
] = [A
1
T
1
]
a
[L]
b
[ML2T
–1
]
c
[LT
–1
]
d
[M
–1
L
–2
T
+4
A
+2
] = [M
c 
L
b+2c+d 
T
a–c–d 
A
a
]
a = 2, b = 1, c = – 1, d = – 1
? u = 
10. (a) = angle.
11. (d) Dimensionally e
0
L = Capacitance (c)
? 
Page 3


1. (a) A and B have different dimensions. Hence, only A and B in a ratio
form and is meaningful.
2. (d) No of divisions on main scale = N
No of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) ? b = 
Least count is a – b = a – 
=  = 
3. (a)
4. (a) T = P
a
 D
b
 S
c
M
0
L
0
T
1
 = (ML
–1
 T
–2
)
a
 (ML
–3
)
b
 (MT
–2
)
c
= M
a+b+c
 L
–a–3b
 T
–2a–2c
Applying principle of homogeneity
a + b + c = 0;  – a – 3b = 0;  – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
5. (b) As 
 
6. (a) Here, Mass of a body, M = 5.00 ± 0.05 kg
Volume of a body, V = 1.00 ± 0.05 m
3
Density, ? = 
Relative error in density is,  
Percentage error in density is
 = 1% + 5% = 6%
7. (d) For angular momentum, the dimensional formula is [ML
2
T
–1
]. For
other three, it is [ML
2
T
–2
].
8. (d) Let X = [ML
–1
 T
–1
]
Then,  
As we know,  
= (1 + 1.5 + 3) % = 5.5 %.
9. (d) Let unit ‘u’ related with e, a
0
, h and c as follows.
[u] = [e]
a
 [a
0
]
b
 [h]
c 
[C]
d
Using dimensional method,
[M
–1
L
–2
T
+4
A
+2
] = [A
1
T
1
]
a
[L]
b
[ML2T
–1
]
c
[LT
–1
]
d
[M
–1
L
–2
T
+4
A
+2
] = [M
c 
L
b+2c+d 
T
a–c–d 
A
a
]
a = 2, b = 1, c = – 1, d = – 1
? u = 
10. (a) = angle.
11. (d) Dimensionally e
0
L = Capacitance (c)
? 
12. (d) P = ,  × 100% =  × 100% + × 100% +  ×
100% +  × 100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
13. (c) We know that  is energy of capacitor so it represent the
dimension of energy = [ML
2
T
–2
].
14. (a) Number of significant figures in 23.023= 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10
–3
 = 2
15. (b) Mobility µ = 
= 
= kg
–1
 s
2
 A = [M
–1
 T
2
 A]
16. (b) [momentum] = [M][L][T
–1
] = [MLT
–1
]
Planck’s constant =  =
17. (b) Measured length of rod = 3.50 cm
For vernier scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
18. (c) Impulse = change in momentum
19. (d)
20. (d) Let dimensions of length is related as,
Page 4


1. (a) A and B have different dimensions. Hence, only A and B in a ratio
form and is meaningful.
2. (d) No of divisions on main scale = N
No of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) ? b = 
Least count is a – b = a – 
=  = 
3. (a)
4. (a) T = P
a
 D
b
 S
c
M
0
L
0
T
1
 = (ML
–1
 T
–2
)
a
 (ML
–3
)
b
 (MT
–2
)
c
= M
a+b+c
 L
–a–3b
 T
–2a–2c
Applying principle of homogeneity
a + b + c = 0;  – a – 3b = 0;  – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
5. (b) As 
 
6. (a) Here, Mass of a body, M = 5.00 ± 0.05 kg
Volume of a body, V = 1.00 ± 0.05 m
3
Density, ? = 
Relative error in density is,  
Percentage error in density is
 = 1% + 5% = 6%
7. (d) For angular momentum, the dimensional formula is [ML
2
T
–1
]. For
other three, it is [ML
2
T
–2
].
8. (d) Let X = [ML
–1
 T
–1
]
Then,  
As we know,  
= (1 + 1.5 + 3) % = 5.5 %.
9. (d) Let unit ‘u’ related with e, a
0
, h and c as follows.
[u] = [e]
a
 [a
0
]
b
 [h]
c 
[C]
d
Using dimensional method,
[M
–1
L
–2
T
+4
A
+2
] = [A
1
T
1
]
a
[L]
b
[ML2T
–1
]
c
[LT
–1
]
d
[M
–1
L
–2
T
+4
A
+2
] = [M
c 
L
b+2c+d 
T
a–c–d 
A
a
]
a = 2, b = 1, c = – 1, d = – 1
? u = 
10. (a) = angle.
11. (d) Dimensionally e
0
L = Capacitance (c)
? 
12. (d) P = ,  × 100% =  × 100% + × 100% +  ×
100% +  × 100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
13. (c) We know that  is energy of capacitor so it represent the
dimension of energy = [ML
2
T
–2
].
14. (a) Number of significant figures in 23.023= 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10
–3
 = 2
15. (b) Mobility µ = 
= 
= kg
–1
 s
2
 A = [M
–1
 T
2
 A]
16. (b) [momentum] = [M][L][T
–1
] = [MLT
–1
]
Planck’s constant =  =
17. (b) Measured length of rod = 3.50 cm
For vernier scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
18. (c) Impulse = change in momentum
19. (d)
20. (d) Let dimensions of length is related as,
L 
  = ML
3
T
–2
L = [LT
–1
]
x
 [M
–1
L
3
T
–2
]
y
[ML
3
T
–2
]
z
[L] = [L
x
 
+ 3y + 3z
 M
 –y + z
 T
–x – 2y – 2z
]
Comparing both sides
–y + z = 0 ? y = z ...(i)
x + 3y + 3z = 1 ...(ii)
–x – 4z = 0 ( Q  y = z) ...(iii)
From (i), (ii) and (iii)
z = y =  x = –2
Hence, L = 
21. (8) As we know, time period of a simple pendulum
T = 2p
The maximum percentage error in g
= 2% + 2(3%) = 8%
22. (3.38) Least count of screw gauge = 
 Reading = [Main scale reading  + circular scale  reading × L.C]
– (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
23. (40) In CGS system,
Page 5


1. (a) A and B have different dimensions. Hence, only A and B in a ratio
form and is meaningful.
2. (d) No of divisions on main scale = N
No of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) ? b = 
Least count is a – b = a – 
=  = 
3. (a)
4. (a) T = P
a
 D
b
 S
c
M
0
L
0
T
1
 = (ML
–1
 T
–2
)
a
 (ML
–3
)
b
 (MT
–2
)
c
= M
a+b+c
 L
–a–3b
 T
–2a–2c
Applying principle of homogeneity
a + b + c = 0;  – a – 3b = 0;  – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
5. (b) As 
 
6. (a) Here, Mass of a body, M = 5.00 ± 0.05 kg
Volume of a body, V = 1.00 ± 0.05 m
3
Density, ? = 
Relative error in density is,  
Percentage error in density is
 = 1% + 5% = 6%
7. (d) For angular momentum, the dimensional formula is [ML
2
T
–1
]. For
other three, it is [ML
2
T
–2
].
8. (d) Let X = [ML
–1
 T
–1
]
Then,  
As we know,  
= (1 + 1.5 + 3) % = 5.5 %.
9. (d) Let unit ‘u’ related with e, a
0
, h and c as follows.
[u] = [e]
a
 [a
0
]
b
 [h]
c 
[C]
d
Using dimensional method,
[M
–1
L
–2
T
+4
A
+2
] = [A
1
T
1
]
a
[L]
b
[ML2T
–1
]
c
[LT
–1
]
d
[M
–1
L
–2
T
+4
A
+2
] = [M
c 
L
b+2c+d 
T
a–c–d 
A
a
]
a = 2, b = 1, c = – 1, d = – 1
? u = 
10. (a) = angle.
11. (d) Dimensionally e
0
L = Capacitance (c)
? 
12. (d) P = ,  × 100% =  × 100% + × 100% +  ×
100% +  × 100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
13. (c) We know that  is energy of capacitor so it represent the
dimension of energy = [ML
2
T
–2
].
14. (a) Number of significant figures in 23.023= 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10
–3
 = 2
15. (b) Mobility µ = 
= 
= kg
–1
 s
2
 A = [M
–1
 T
2
 A]
16. (b) [momentum] = [M][L][T
–1
] = [MLT
–1
]
Planck’s constant =  =
17. (b) Measured length of rod = 3.50 cm
For vernier scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
18. (c) Impulse = change in momentum
19. (d)
20. (d) Let dimensions of length is related as,
L 
  = ML
3
T
–2
L = [LT
–1
]
x
 [M
–1
L
3
T
–2
]
y
[ML
3
T
–2
]
z
[L] = [L
x
 
+ 3y + 3z
 M
 –y + z
 T
–x – 2y – 2z
]
Comparing both sides
–y + z = 0 ? y = z ...(i)
x + 3y + 3z = 1 ...(ii)
–x – 4z = 0 ( Q  y = z) ...(iii)
From (i), (ii) and (iii)
z = y =  x = –2
Hence, L = 
21. (8) As we know, time period of a simple pendulum
T = 2p
The maximum percentage error in g
= 2% + 2(3%) = 8%
22. (3.38) Least count of screw gauge = 
 Reading = [Main scale reading  + circular scale  reading × L.C]
– (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
23. (40) In CGS system,
The unit of mass is 100g and unit of length is 10 cm, so
density = 
= 
= 40 unit
24. (0.2)The current voltage relation of diode is
I = (e
1000 V/T
 – 1) mA (given)
When, I = 5 mA, e
1000 V/T
 = 6mA
Also,  (By exponential function)
=  = 0.2 mA
25. (3.1)Diameter D = M.S.R. + (C.S.R) × L.C.
D = 2.5 + 20 × 
D = 2.70 mm
The uncertainty in the measurement of diameter?D = 0.01 mm.
We know that
? =  =  = 
 =

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FAQs on Physical World, Units & Measurements Practice Questions - DPP for JEE

1. What are the basic concepts of units and measurements in physics?

Ans. The basic concepts of units and measurements in physics revolve around quantifying physical quantities. A unit is a standard measure of a quantity (e.g., meter for length, kilogram for mass). Measurements involve comparing an unknown quantity to a standard unit. The International System of Units (SI) is commonly used, which includes base units like meter, kilogram, second, ampere, kelvin, mole, and candela. Understanding these concepts is fundamental for solving problems in physics effectively.

2. How do you convert between different units in measurements?

Ans. Converting between different units involves using conversion factors, which are ratios that express how many of one unit are equivalent to another. For example, to convert meters to kilometers, you use the factor \(1 \text{ km} = 1000 \text{ m}\). You multiply the value in meters by the conversion factor (1 km/1000 m) to obtain the equivalent value in kilometers. It's essential to ensure that the units cancel appropriately during the conversion process.

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

Ans. Significant figures are important in measurements as they indicate the precision of a measurement. They include all the digits that are known with certainty, plus one last digit that is estimated. The number of significant figures reflects the accuracy of the measuring instrument and affects calculations. When performing calculations, the result should be reported with the appropriate number of significant figures to communicate the level of uncertainty in the measurement accurately.

4. What are systematic and random errors in measurements?

Ans. Systematic errors are consistent, repeatable errors associated with faulty equipment or measurement techniques, leading to inaccurate results. For instance, a scale that is not calibrated properly will always give the same incorrect reading. Random errors, on the other hand, fluctuate unpredictably due to environmental conditions or human mistakes, resulting in varying measurements. Understanding these errors is crucial for improving accuracy and reliability in experimental results.

5. How can I improve my skills in units and measurements for JEE preparation?

Ans. To improve skills in units and measurements for JEE preparation, practice is key. Start by reviewing the fundamental concepts and definitions, then solve a variety of numerical problems to apply these concepts. Use study materials like previous years' JEE papers, online quizzes, and mock tests to familiarize yourself with the exam format. Additionally, consider joining coaching classes or study groups to clarify doubts and gain different perspectives on problem-solving techniques.

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