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

 Page 1


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, ? = 
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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, ? = 
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)
? 
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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, ? = 
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
? =  =  = 
 =