QBManager

Units and Measurements

139847 The velocity of projection of a body is increased by $2 %$ . Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

1

1 %

2

2 %

3

4 %

4

8 %

Explanation:

C Height of projection
$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow H \propto u^{2}, (\frac{\sin^{2} θ}{2 g}) = constant$
taking $\log$ and differentiating
$\log H = 2 \log 4$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u}$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u} \Rightarrow \frac{Δ H}{H} \times 100 = 2 \frac{Δ u}{u} \times 100$
$= 2 (2 %) = 4 %$

AIIMS-25.05.2019(E) Shift-2

Units and Measurements

139848 A wire has a mass $(0.1 \pm 0.001) g$ radius $(0.5 \pm$ $0.005) mm$ and length $(10 \pm 0.1) cm$ . The maximum percentage error in the measurement of its density is.

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

D Density $D = \frac{mass (m)}{volume (v)}$
$\log D = \log m + \log v {∵$ for maximum $}$
$\frac{Δ D}{D} = \frac{Δ M}{M} = \frac{Δ V}{V}$
$\frac{Δ D}{D} \times 100 = \frac{Δ M}{M} \times 100 = \frac{Δ V}{V} \times 100$
$= \frac{0.001}{0.1} \times 100 + (2 \times \frac{0.005}{0.5} + \frac{0.1}{10}) \times 100$
$= 0.01 + 0.02 + 0.01 \times 100$
$= 0.04 \times 100 = 4 %$

UPSEE 2019

Units and Measurements

139849 A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 cm$ . The main scale reading is $5 mm$ and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of $- 0.004 cm$ , the correct diameter of the ball is

1

0.053 cm

2

0.525 cm

3

0.521 cm

4

0.529 cm

Explanation:

C Given that,
Main scale reading $(MSR) = 5 mm = 0.5 cm$
Circular scale division $(CSD) = 25$
least count $= 0.001 cm$
zero error $= - 0.004 cm$
Thus,
Diameter $=$ calculated diameter value - zero error
$= MSR + CSD \times LC - zero error$
$= 0.5 + 25 \times 0.001 - (- 0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529 cm$

NEET 2018

Units and Measurements

139850 The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5 %$ and $1 %$ , the maximum error in determining the density is

1

2.5 %

2

3.5 %

3

4.5 %

4

6 %

Explanation:

C Given that,
$%$ Error in mass $(\frac{Δ m}{m} \times 100) = 1.5 %$
$%$ Error in length $(\frac{Δ l}{l} \times 100) = 1 %$
Density $= \frac{m}{v}$
Volume of cube $= l^{3}$
So,
Relative Error in density $(ρ) = \frac{m}{l^{3}}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 3 \frac{Δ l}{l}) \times 100$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 \times 1 %$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 %$
$\frac{Δ ρ}{ρ} \times 100 = 4.5 %$

JEE Main 2018

Units and Measurements

139847 The velocity of projection of a body is increased by $2 %$ . Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

1

1 %

2

2 %

3

4 %

4

8 %

Explanation:

C Height of projection
$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow H \propto u^{2}, (\frac{\sin^{2} θ}{2 g}) = constant$
taking $\log$ and differentiating
$\log H = 2 \log 4$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u}$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u} \Rightarrow \frac{Δ H}{H} \times 100 = 2 \frac{Δ u}{u} \times 100$
$= 2 (2 %) = 4 %$

AIIMS-25.05.2019(E) Shift-2

Units and Measurements

139848 A wire has a mass $(0.1 \pm 0.001) g$ radius $(0.5 \pm$ $0.005) mm$ and length $(10 \pm 0.1) cm$ . The maximum percentage error in the measurement of its density is.

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

D Density $D = \frac{mass (m)}{volume (v)}$
$\log D = \log m + \log v {∵$ for maximum $}$
$\frac{Δ D}{D} = \frac{Δ M}{M} = \frac{Δ V}{V}$
$\frac{Δ D}{D} \times 100 = \frac{Δ M}{M} \times 100 = \frac{Δ V}{V} \times 100$
$= \frac{0.001}{0.1} \times 100 + (2 \times \frac{0.005}{0.5} + \frac{0.1}{10}) \times 100$
$= 0.01 + 0.02 + 0.01 \times 100$
$= 0.04 \times 100 = 4 %$

UPSEE 2019

Units and Measurements

139849 A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 cm$ . The main scale reading is $5 mm$ and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of $- 0.004 cm$ , the correct diameter of the ball is

1

0.053 cm

2

0.525 cm

3

0.521 cm

4

0.529 cm

Explanation:

C Given that,
Main scale reading $(MSR) = 5 mm = 0.5 cm$
Circular scale division $(CSD) = 25$
least count $= 0.001 cm$
zero error $= - 0.004 cm$
Thus,
Diameter $=$ calculated diameter value - zero error
$= MSR + CSD \times LC - zero error$
$= 0.5 + 25 \times 0.001 - (- 0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529 cm$

NEET 2018

Units and Measurements

139850 The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5 %$ and $1 %$ , the maximum error in determining the density is

1

2.5 %

2

3.5 %

3

4.5 %

4

6 %

Explanation:

C Given that,
$%$ Error in mass $(\frac{Δ m}{m} \times 100) = 1.5 %$
$%$ Error in length $(\frac{Δ l}{l} \times 100) = 1 %$
Density $= \frac{m}{v}$
Volume of cube $= l^{3}$
So,
Relative Error in density $(ρ) = \frac{m}{l^{3}}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 3 \frac{Δ l}{l}) \times 100$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 \times 1 %$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 %$
$\frac{Δ ρ}{ρ} \times 100 = 4.5 %$

JEE Main 2018

Units and Measurements

139847 The velocity of projection of a body is increased by $2 %$ . Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

1

1 %

2

2 %

3

4 %

4

8 %

Explanation:

C Height of projection
$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow H \propto u^{2}, (\frac{\sin^{2} θ}{2 g}) = constant$
taking $\log$ and differentiating
$\log H = 2 \log 4$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u}$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u} \Rightarrow \frac{Δ H}{H} \times 100 = 2 \frac{Δ u}{u} \times 100$
$= 2 (2 %) = 4 %$

AIIMS-25.05.2019(E) Shift-2

Units and Measurements

139848 A wire has a mass $(0.1 \pm 0.001) g$ radius $(0.5 \pm$ $0.005) mm$ and length $(10 \pm 0.1) cm$ . The maximum percentage error in the measurement of its density is.

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

D Density $D = \frac{mass (m)}{volume (v)}$
$\log D = \log m + \log v {∵$ for maximum $}$
$\frac{Δ D}{D} = \frac{Δ M}{M} = \frac{Δ V}{V}$
$\frac{Δ D}{D} \times 100 = \frac{Δ M}{M} \times 100 = \frac{Δ V}{V} \times 100$
$= \frac{0.001}{0.1} \times 100 + (2 \times \frac{0.005}{0.5} + \frac{0.1}{10}) \times 100$
$= 0.01 + 0.02 + 0.01 \times 100$
$= 0.04 \times 100 = 4 %$

UPSEE 2019

Units and Measurements

139849 A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 cm$ . The main scale reading is $5 mm$ and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of $- 0.004 cm$ , the correct diameter of the ball is

1

0.053 cm

2

0.525 cm

3

0.521 cm

4

0.529 cm

Explanation:

C Given that,
Main scale reading $(MSR) = 5 mm = 0.5 cm$
Circular scale division $(CSD) = 25$
least count $= 0.001 cm$
zero error $= - 0.004 cm$
Thus,
Diameter $=$ calculated diameter value - zero error
$= MSR + CSD \times LC - zero error$
$= 0.5 + 25 \times 0.001 - (- 0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529 cm$

NEET 2018

Units and Measurements

139850 The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5 %$ and $1 %$ , the maximum error in determining the density is

1

2.5 %

2

3.5 %

3

4.5 %

4

6 %

Explanation:

C Given that,
$%$ Error in mass $(\frac{Δ m}{m} \times 100) = 1.5 %$
$%$ Error in length $(\frac{Δ l}{l} \times 100) = 1 %$
Density $= \frac{m}{v}$
Volume of cube $= l^{3}$
So,
Relative Error in density $(ρ) = \frac{m}{l^{3}}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 3 \frac{Δ l}{l}) \times 100$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 \times 1 %$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 %$
$\frac{Δ ρ}{ρ} \times 100 = 4.5 %$

JEE Main 2018

Units and Measurements

139847 The velocity of projection of a body is increased by $2 %$ . Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

1

1 %

2

2 %

3

4 %

4

8 %

Explanation:

C Height of projection
$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow H \propto u^{2}, (\frac{\sin^{2} θ}{2 g}) = constant$
taking $\log$ and differentiating
$\log H = 2 \log 4$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u}$
$\frac{Δ H}{H} = 2 \frac{Δ u}{u} \Rightarrow \frac{Δ H}{H} \times 100 = 2 \frac{Δ u}{u} \times 100$
$= 2 (2 %) = 4 %$

AIIMS-25.05.2019(E) Shift-2

Units and Measurements

139848 A wire has a mass $(0.1 \pm 0.001) g$ radius $(0.5 \pm$ $0.005) mm$ and length $(10 \pm 0.1) cm$ . The maximum percentage error in the measurement of its density is.

1

1 %

2

2 %

3

3 %

4

4 %

Explanation:

D Density $D = \frac{mass (m)}{volume (v)}$
$\log D = \log m + \log v {∵$ for maximum $}$
$\frac{Δ D}{D} = \frac{Δ M}{M} = \frac{Δ V}{V}$
$\frac{Δ D}{D} \times 100 = \frac{Δ M}{M} \times 100 = \frac{Δ V}{V} \times 100$
$= \frac{0.001}{0.1} \times 100 + (2 \times \frac{0.005}{0.5} + \frac{0.1}{10}) \times 100$
$= 0.01 + 0.02 + 0.01 \times 100$
$= 0.04 \times 100 = 4 %$

UPSEE 2019

Units and Measurements

139849 A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 cm$ . The main scale reading is $5 mm$ and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of $- 0.004 cm$ , the correct diameter of the ball is

1

0.053 cm

2

0.525 cm

3

0.521 cm

4

0.529 cm

Explanation:

C Given that,
Main scale reading $(MSR) = 5 mm = 0.5 cm$
Circular scale division $(CSD) = 25$
least count $= 0.001 cm$
zero error $= - 0.004 cm$
Thus,
Diameter $=$ calculated diameter value - zero error
$= MSR + CSD \times LC - zero error$
$= 0.5 + 25 \times 0.001 - (- 0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529 cm$

NEET 2018

Units and Measurements

139850 The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5 %$ and $1 %$ , the maximum error in determining the density is

1

2.5 %

2

3.5 %

3

4.5 %

4

6 %

Explanation:

C Given that,
$%$ Error in mass $(\frac{Δ m}{m} \times 100) = 1.5 %$
$%$ Error in length $(\frac{Δ l}{l} \times 100) = 1 %$
Density $= \frac{m}{v}$
Volume of cube $= l^{3}$
So,
Relative Error in density $(ρ) = \frac{m}{l^{3}}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ m}{m} + 3 \frac{Δ l}{l}) \times 100$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 \times 1 %$
$\frac{Δ ρ}{ρ} \times 100 = 1.5 % + 3 %$
$\frac{Δ ρ}{ρ} \times 100 = 4.5 %$

JEE Main 2018

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: