QBManager

Units and Measurements

139399 The angle of $1^{'}$ (minute of arc) in radian is nearly equal to

1

2.91 \times 10^{- 4} rad

2

4.85 \times 10^{- 4} rad

3

4.80 \times 10^{- 6} rad

4

1.75 \times 10^{- 2} rad

Explanation:

A We know that,
$1^{'} ($ minute of arc $) = (\frac{1}{60})$
$= \frac{1}{60} \times \frac{π}{180}$ radian
$= \frac{π}{60 \times 180}$
$= \frac{3.14}{10800}$
$= 0.0291 \times 10^{- 2}$
$= 2.91 \times 10^{- 4}$ radian

NEET (Oct.) 2020

Units and Measurements

139401 The density of a material in SI units is $128 kg$ $m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

1 40

2 16

3 640

4 410

Explanation:

A We know that
Density $= \frac{Mass}{Volume} = \frac{M}{V} = \frac{M}{L^{3}}$
Then, $n_{1} u_{1} = n_{2} u_{2}$
$128 [M_{1} L_{1}^{- 3}] = n_{2} [M_{2} L_{2}^{- 3}]$
$n_{2} = \frac{128 [M_{1} L_{1}^{- 3}]}{[M_{2} L_{2}^{- 3}]}$
$n_{1} =$ numerical value in S.I unit
$u_{1} =$ unit in S.I.
$n_{2} =$ numerical value in other unit
$u_{2} =$ unit in other system
In second unit system $m = 50 g = \frac{50}{1000} kg$ .
$L = 25 cm = \frac{25}{100} mtr$
$n_{2} = 128 [\frac{M_{1}}{M_{2}}] {[\frac{L_{2}}{L_{1}}]}^{3}$
$= 128 \times [\frac{1}{\frac{50}{1000}}] \times {[\frac{\frac{25}{100}}{1}]}^{3}$
$= 128 \times 20 \times {(\frac{1}{4})}^{3}$
$= 128 \times 20 \times (\frac{1}{64})$
$= 40$

JEE Main-10.01.2019

Units and Measurements

139404 What is dimensions of energy in terms of linear momentum (P), area of $(A)$ and Time (T)

1

[P^{1} A^{1} T^{1}]

2

[P^{2} A^{2} T^{- 1}]

3

[P^{1} A^{1 / 2} T^{- 1}]

4

[P^{1 / 2} A^{1 / 2} T^{- 1 -}]

Explanation:

C Let,
$Energy E = {kP}^{a} A^{b} T^{c}$
Where $k$ is a dimensionless constant of proportionalityWriting dimension on both sides-
$[{ML}^{2} T^{- 2}] = {[{MLT}^{- 1}]}^{a} {[M^{0} L^{2} T^{0}]}^{b} {[M^{0} L^{0} T]}^{c}$
$[{ML}^{2} T^{- 2}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the power from both sides-
$a = 1$
$a + 2 b = 2$
$- a + c = - 2$
On solving (ii), (iii) and (iv), we have-
$\begin{array}{ll} a = 1, b = 1 / 2, c = \\ So, [E] = [P^{1} A^{1 / 2} T^{- 1}] \end{array}$

JIPMER-2019

Units and Measurements

139405 If $10 g {cms}^{- 1} = x$ Ns. then the number is

1

1 \times 10^{- 5}

2

1 \times 10^{- 4}

3

1 \times 10^{- 6}

4

1 \times 10^{- 3}

Explanation:

A We know that
$1 g = 10^{- 3} kg$
$1 cm = 10^{- 2} m$
According to question-
$10 {gcms}^{- 1} = 10^{- 3} kg \times 10^{- 2} m \times \sec^{- 1}$
$= 10^{- 5} kg m \cdot \sec^{- 1}$
$= 10^{- 5} Ns [N = \frac{kgm}{\sec}]$

AP EAMCET-24.04.2018

Units and Measurements

139407 A star is very far from earth. If light 10 years from it to reach the earth, calculate the distance between star and earth.

1

9.45 \times 10^{16} m

2

9.46 \times 10^{- 16} m

3

9.46 \times 10^{17} m

4

9.46 \times 10^{- 17} m

[SRM JEE-2018]

Explanation:

A We know that One-light year is the distance Travelled by light in one year.
So, light year $= 9.46 \times 10^{12} km = 9.46 \times 10^{15} m$
So,
$10 light year = 9.4 \times 10^{15} \times 10 m$
$= 9.46 \times 10^{16} m$

Units and Measurements

139399 The angle of $1^{'}$ (minute of arc) in radian is nearly equal to

1

2.91 \times 10^{- 4} rad

2

4.85 \times 10^{- 4} rad

3

4.80 \times 10^{- 6} rad

4

1.75 \times 10^{- 2} rad

Explanation:

A We know that,
$1^{'} ($ minute of arc $) = (\frac{1}{60})$
$= \frac{1}{60} \times \frac{π}{180}$ radian
$= \frac{π}{60 \times 180}$
$= \frac{3.14}{10800}$
$= 0.0291 \times 10^{- 2}$
$= 2.91 \times 10^{- 4}$ radian

NEET (Oct.) 2020

Units and Measurements

139401 The density of a material in SI units is $128 kg$ $m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

1 40

2 16

3 640

4 410

Explanation:

A We know that
Density $= \frac{Mass}{Volume} = \frac{M}{V} = \frac{M}{L^{3}}$
Then, $n_{1} u_{1} = n_{2} u_{2}$
$128 [M_{1} L_{1}^{- 3}] = n_{2} [M_{2} L_{2}^{- 3}]$
$n_{2} = \frac{128 [M_{1} L_{1}^{- 3}]}{[M_{2} L_{2}^{- 3}]}$
$n_{1} =$ numerical value in S.I unit
$u_{1} =$ unit in S.I.
$n_{2} =$ numerical value in other unit
$u_{2} =$ unit in other system
In second unit system $m = 50 g = \frac{50}{1000} kg$ .
$L = 25 cm = \frac{25}{100} mtr$
$n_{2} = 128 [\frac{M_{1}}{M_{2}}] {[\frac{L_{2}}{L_{1}}]}^{3}$
$= 128 \times [\frac{1}{\frac{50}{1000}}] \times {[\frac{\frac{25}{100}}{1}]}^{3}$
$= 128 \times 20 \times {(\frac{1}{4})}^{3}$
$= 128 \times 20 \times (\frac{1}{64})$
$= 40$

JEE Main-10.01.2019

Units and Measurements

139404 What is dimensions of energy in terms of linear momentum (P), area of $(A)$ and Time (T)

1

[P^{1} A^{1} T^{1}]

2

[P^{2} A^{2} T^{- 1}]

3

[P^{1} A^{1 / 2} T^{- 1}]

4

[P^{1 / 2} A^{1 / 2} T^{- 1 -}]

Explanation:

C Let,
$Energy E = {kP}^{a} A^{b} T^{c}$
Where $k$ is a dimensionless constant of proportionalityWriting dimension on both sides-
$[{ML}^{2} T^{- 2}] = {[{MLT}^{- 1}]}^{a} {[M^{0} L^{2} T^{0}]}^{b} {[M^{0} L^{0} T]}^{c}$
$[{ML}^{2} T^{- 2}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the power from both sides-
$a = 1$
$a + 2 b = 2$
$- a + c = - 2$
On solving (ii), (iii) and (iv), we have-
$\begin{array}{ll} a = 1, b = 1 / 2, c = \\ So, [E] = [P^{1} A^{1 / 2} T^{- 1}] \end{array}$

JIPMER-2019

Units and Measurements

139405 If $10 g {cms}^{- 1} = x$ Ns. then the number is

1

1 \times 10^{- 5}

2

1 \times 10^{- 4}

3

1 \times 10^{- 6}

4

1 \times 10^{- 3}

Explanation:

A We know that
$1 g = 10^{- 3} kg$
$1 cm = 10^{- 2} m$
According to question-
$10 {gcms}^{- 1} = 10^{- 3} kg \times 10^{- 2} m \times \sec^{- 1}$
$= 10^{- 5} kg m \cdot \sec^{- 1}$
$= 10^{- 5} Ns [N = \frac{kgm}{\sec}]$

AP EAMCET-24.04.2018

Units and Measurements

139407 A star is very far from earth. If light 10 years from it to reach the earth, calculate the distance between star and earth.

1

9.45 \times 10^{16} m

2

9.46 \times 10^{- 16} m

3

9.46 \times 10^{17} m

4

9.46 \times 10^{- 17} m

[SRM JEE-2018]

Explanation:

A We know that One-light year is the distance Travelled by light in one year.
So, light year $= 9.46 \times 10^{12} km = 9.46 \times 10^{15} m$
So,
$10 light year = 9.4 \times 10^{15} \times 10 m$
$= 9.46 \times 10^{16} m$

Units and Measurements

139399 The angle of $1^{'}$ (minute of arc) in radian is nearly equal to

1

2.91 \times 10^{- 4} rad

2

4.85 \times 10^{- 4} rad

3

4.80 \times 10^{- 6} rad

4

1.75 \times 10^{- 2} rad

Explanation:

A We know that,
$1^{'} ($ minute of arc $) = (\frac{1}{60})$
$= \frac{1}{60} \times \frac{π}{180}$ radian
$= \frac{π}{60 \times 180}$
$= \frac{3.14}{10800}$
$= 0.0291 \times 10^{- 2}$
$= 2.91 \times 10^{- 4}$ radian

NEET (Oct.) 2020

Units and Measurements

139401 The density of a material in SI units is $128 kg$ $m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

1 40

2 16

3 640

4 410

Explanation:

A We know that
Density $= \frac{Mass}{Volume} = \frac{M}{V} = \frac{M}{L^{3}}$
Then, $n_{1} u_{1} = n_{2} u_{2}$
$128 [M_{1} L_{1}^{- 3}] = n_{2} [M_{2} L_{2}^{- 3}]$
$n_{2} = \frac{128 [M_{1} L_{1}^{- 3}]}{[M_{2} L_{2}^{- 3}]}$
$n_{1} =$ numerical value in S.I unit
$u_{1} =$ unit in S.I.
$n_{2} =$ numerical value in other unit
$u_{2} =$ unit in other system
In second unit system $m = 50 g = \frac{50}{1000} kg$ .
$L = 25 cm = \frac{25}{100} mtr$
$n_{2} = 128 [\frac{M_{1}}{M_{2}}] {[\frac{L_{2}}{L_{1}}]}^{3}$
$= 128 \times [\frac{1}{\frac{50}{1000}}] \times {[\frac{\frac{25}{100}}{1}]}^{3}$
$= 128 \times 20 \times {(\frac{1}{4})}^{3}$
$= 128 \times 20 \times (\frac{1}{64})$
$= 40$

JEE Main-10.01.2019

Units and Measurements

139404 What is dimensions of energy in terms of linear momentum (P), area of $(A)$ and Time (T)

1

[P^{1} A^{1} T^{1}]

2

[P^{2} A^{2} T^{- 1}]

3

[P^{1} A^{1 / 2} T^{- 1}]

4

[P^{1 / 2} A^{1 / 2} T^{- 1 -}]

Explanation:

C Let,
$Energy E = {kP}^{a} A^{b} T^{c}$
Where $k$ is a dimensionless constant of proportionalityWriting dimension on both sides-
$[{ML}^{2} T^{- 2}] = {[{MLT}^{- 1}]}^{a} {[M^{0} L^{2} T^{0}]}^{b} {[M^{0} L^{0} T]}^{c}$
$[{ML}^{2} T^{- 2}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the power from both sides-
$a = 1$
$a + 2 b = 2$
$- a + c = - 2$
On solving (ii), (iii) and (iv), we have-
$\begin{array}{ll} a = 1, b = 1 / 2, c = \\ So, [E] = [P^{1} A^{1 / 2} T^{- 1}] \end{array}$

JIPMER-2019

Units and Measurements

139405 If $10 g {cms}^{- 1} = x$ Ns. then the number is

1

1 \times 10^{- 5}

2

1 \times 10^{- 4}

3

1 \times 10^{- 6}

4

1 \times 10^{- 3}

Explanation:

A We know that
$1 g = 10^{- 3} kg$
$1 cm = 10^{- 2} m$
According to question-
$10 {gcms}^{- 1} = 10^{- 3} kg \times 10^{- 2} m \times \sec^{- 1}$
$= 10^{- 5} kg m \cdot \sec^{- 1}$
$= 10^{- 5} Ns [N = \frac{kgm}{\sec}]$

AP EAMCET-24.04.2018

Units and Measurements

139407 A star is very far from earth. If light 10 years from it to reach the earth, calculate the distance between star and earth.

1

9.45 \times 10^{16} m

2

9.46 \times 10^{- 16} m

3

9.46 \times 10^{17} m

4

9.46 \times 10^{- 17} m

[SRM JEE-2018]

Explanation:

A We know that One-light year is the distance Travelled by light in one year.
So, light year $= 9.46 \times 10^{12} km = 9.46 \times 10^{15} m$
So,
$10 light year = 9.4 \times 10^{15} \times 10 m$
$= 9.46 \times 10^{16} m$

Units and Measurements

139399 The angle of $1^{'}$ (minute of arc) in radian is nearly equal to

1

2.91 \times 10^{- 4} rad

2

4.85 \times 10^{- 4} rad

3

4.80 \times 10^{- 6} rad

4

1.75 \times 10^{- 2} rad

Explanation:

A We know that,
$1^{'} ($ minute of arc $) = (\frac{1}{60})$
$= \frac{1}{60} \times \frac{π}{180}$ radian
$= \frac{π}{60 \times 180}$
$= \frac{3.14}{10800}$
$= 0.0291 \times 10^{- 2}$
$= 2.91 \times 10^{- 4}$ radian

NEET (Oct.) 2020

Units and Measurements

139401 The density of a material in SI units is $128 kg$ $m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

1 40

2 16

3 640

4 410

Explanation:

A We know that
Density $= \frac{Mass}{Volume} = \frac{M}{V} = \frac{M}{L^{3}}$
Then, $n_{1} u_{1} = n_{2} u_{2}$
$128 [M_{1} L_{1}^{- 3}] = n_{2} [M_{2} L_{2}^{- 3}]$
$n_{2} = \frac{128 [M_{1} L_{1}^{- 3}]}{[M_{2} L_{2}^{- 3}]}$
$n_{1} =$ numerical value in S.I unit
$u_{1} =$ unit in S.I.
$n_{2} =$ numerical value in other unit
$u_{2} =$ unit in other system
In second unit system $m = 50 g = \frac{50}{1000} kg$ .
$L = 25 cm = \frac{25}{100} mtr$
$n_{2} = 128 [\frac{M_{1}}{M_{2}}] {[\frac{L_{2}}{L_{1}}]}^{3}$
$= 128 \times [\frac{1}{\frac{50}{1000}}] \times {[\frac{\frac{25}{100}}{1}]}^{3}$
$= 128 \times 20 \times {(\frac{1}{4})}^{3}$
$= 128 \times 20 \times (\frac{1}{64})$
$= 40$

JEE Main-10.01.2019

Units and Measurements

139404 What is dimensions of energy in terms of linear momentum (P), area of $(A)$ and Time (T)

1

[P^{1} A^{1} T^{1}]

2

[P^{2} A^{2} T^{- 1}]

3

[P^{1} A^{1 / 2} T^{- 1}]

4

[P^{1 / 2} A^{1 / 2} T^{- 1 -}]

Explanation:

C Let,
$Energy E = {kP}^{a} A^{b} T^{c}$
Where $k$ is a dimensionless constant of proportionalityWriting dimension on both sides-
$[{ML}^{2} T^{- 2}] = {[{MLT}^{- 1}]}^{a} {[M^{0} L^{2} T^{0}]}^{b} {[M^{0} L^{0} T]}^{c}$
$[{ML}^{2} T^{- 2}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the power from both sides-
$a = 1$
$a + 2 b = 2$
$- a + c = - 2$
On solving (ii), (iii) and (iv), we have-
$\begin{array}{ll} a = 1, b = 1 / 2, c = \\ So, [E] = [P^{1} A^{1 / 2} T^{- 1}] \end{array}$

JIPMER-2019

Units and Measurements

139405 If $10 g {cms}^{- 1} = x$ Ns. then the number is

1

1 \times 10^{- 5}

2

1 \times 10^{- 4}

3

1 \times 10^{- 6}

4

1 \times 10^{- 3}

Explanation:

A We know that
$1 g = 10^{- 3} kg$
$1 cm = 10^{- 2} m$
According to question-
$10 {gcms}^{- 1} = 10^{- 3} kg \times 10^{- 2} m \times \sec^{- 1}$
$= 10^{- 5} kg m \cdot \sec^{- 1}$
$= 10^{- 5} Ns [N = \frac{kgm}{\sec}]$

AP EAMCET-24.04.2018

Units and Measurements

139407 A star is very far from earth. If light 10 years from it to reach the earth, calculate the distance between star and earth.

1

9.45 \times 10^{16} m

2

9.46 \times 10^{- 16} m

3

9.46 \times 10^{17} m

4

9.46 \times 10^{- 17} m

[SRM JEE-2018]

Explanation:

A We know that One-light year is the distance Travelled by light in one year.
So, light year $= 9.46 \times 10^{12} km = 9.46 \times 10^{15} m$
So,
$10 light year = 9.4 \times 10^{15} \times 10 m$
$= 9.46 \times 10^{16} m$

Units and Measurements

139399 The angle of $1^{'}$ (minute of arc) in radian is nearly equal to

1

2.91 \times 10^{- 4} rad

2

4.85 \times 10^{- 4} rad

3

4.80 \times 10^{- 6} rad

4

1.75 \times 10^{- 2} rad

Explanation:

A We know that,
$1^{'} ($ minute of arc $) = (\frac{1}{60})$
$= \frac{1}{60} \times \frac{π}{180}$ radian
$= \frac{π}{60 \times 180}$
$= \frac{3.14}{10800}$
$= 0.0291 \times 10^{- 2}$
$= 2.91 \times 10^{- 4}$ radian

NEET (Oct.) 2020

Units and Measurements

139401 The density of a material in SI units is $128 kg$ $m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

1 40

2 16

3 640

4 410

Explanation:

A We know that
Density $= \frac{Mass}{Volume} = \frac{M}{V} = \frac{M}{L^{3}}$
Then, $n_{1} u_{1} = n_{2} u_{2}$
$128 [M_{1} L_{1}^{- 3}] = n_{2} [M_{2} L_{2}^{- 3}]$
$n_{2} = \frac{128 [M_{1} L_{1}^{- 3}]}{[M_{2} L_{2}^{- 3}]}$
$n_{1} =$ numerical value in S.I unit
$u_{1} =$ unit in S.I.
$n_{2} =$ numerical value in other unit
$u_{2} =$ unit in other system
In second unit system $m = 50 g = \frac{50}{1000} kg$ .
$L = 25 cm = \frac{25}{100} mtr$
$n_{2} = 128 [\frac{M_{1}}{M_{2}}] {[\frac{L_{2}}{L_{1}}]}^{3}$
$= 128 \times [\frac{1}{\frac{50}{1000}}] \times {[\frac{\frac{25}{100}}{1}]}^{3}$
$= 128 \times 20 \times {(\frac{1}{4})}^{3}$
$= 128 \times 20 \times (\frac{1}{64})$
$= 40$

JEE Main-10.01.2019

Units and Measurements

139404 What is dimensions of energy in terms of linear momentum (P), area of $(A)$ and Time (T)

1

[P^{1} A^{1} T^{1}]

2

[P^{2} A^{2} T^{- 1}]

3

[P^{1} A^{1 / 2} T^{- 1}]

4

[P^{1 / 2} A^{1 / 2} T^{- 1 -}]

Explanation:

C Let,
$Energy E = {kP}^{a} A^{b} T^{c}$
Where $k$ is a dimensionless constant of proportionalityWriting dimension on both sides-
$[{ML}^{2} T^{- 2}] = {[{MLT}^{- 1}]}^{a} {[M^{0} L^{2} T^{0}]}^{b} {[M^{0} L^{0} T]}^{c}$
$[{ML}^{2} T^{- 2}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the power from both sides-
$a = 1$
$a + 2 b = 2$
$- a + c = - 2$
On solving (ii), (iii) and (iv), we have-
$\begin{array}{ll} a = 1, b = 1 / 2, c = \\ So, [E] = [P^{1} A^{1 / 2} T^{- 1}] \end{array}$

JIPMER-2019

Units and Measurements

139405 If $10 g {cms}^{- 1} = x$ Ns. then the number is

1

1 \times 10^{- 5}

2

1 \times 10^{- 4}

3

1 \times 10^{- 6}

4

1 \times 10^{- 3}

Explanation:

A We know that
$1 g = 10^{- 3} kg$
$1 cm = 10^{- 2} m$
According to question-
$10 {gcms}^{- 1} = 10^{- 3} kg \times 10^{- 2} m \times \sec^{- 1}$
$= 10^{- 5} kg m \cdot \sec^{- 1}$
$= 10^{- 5} Ns [N = \frac{kgm}{\sec}]$

AP EAMCET-24.04.2018

Units and Measurements

139407 A star is very far from earth. If light 10 years from it to reach the earth, calculate the distance between star and earth.

1

9.45 \times 10^{16} m

2

9.46 \times 10^{- 16} m

3

9.46 \times 10^{17} m

4

9.46 \times 10^{- 17} m

[SRM JEE-2018]

Explanation:

A We know that One-light year is the distance Travelled by light in one year.
So, light year $= 9.46 \times 10^{12} km = 9.46 \times 10^{15} m$
So,
$10 light year = 9.4 \times 10^{15} \times 10 m$
$= 9.46 \times 10^{16} m$