QBManager

Units and Measurements

139408 It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun. This is called solar constant, the value of solar constant in S.I. units is

1

2 J m^{2} s^{- 1}

2

1.4 {Wm}^{- 2}

3

2.4 {kWm}^{- 2}

4

1.4 {kJm}^{- 2} s^{- 1}

Explanation:

D Given that,
It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun.
$S = 2 cal / {cm}^{2} - \min$
We know that,
$1 cal = 4.18 J$
$1 cm = 10^{- 2} m$
$1 \min = 60 \sec$
$∴ S = \frac{2 \times 4.18}{{(10^{- 2})}^{2} \times 60} \frac{J}{m^{2} - \sec}$
$S = \frac{2 \times 4.18}{10^{- 4} \times 60}$
$S = \frac{4.18 \times 10^{4}}{30}$
$S = 0.13933 \times 10^{4}$
$S = 0.14 \times 10^{4}$
$S = 1.4 \times 10^{3} \frac{J}{m^{2} - \sec}$
$S = 1.4 \frac{kJ}{m^{2} - \sec}$
Putting the value of $z$ in ${eq}^{n}$ .
$- x - 2 y - 2 z = 0$
$- x - 2 y - 2 y = 0$
$- x - 4 y = 0$
$x = - 4 y$
Now putting the value of $x$ in $e^{n}$
$x + 6 y = 1$
$- 4 y + 6 y = 1$
$2 y = 1$
$y = \frac{1}{2}$
from eq $^{n}$ (i)
$z = y$

AP EAMCET-25.04.2017

Units and Measurements

139409 A physical quantity of the dimensions of length that can be formed out of $C, G$ and $\frac{e^{2}}{4 π ε_{0}}$ is [c is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]

1

\frac{1}{C^{2}} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

2

C^{2} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

3

\frac{1}{C^{2}} [G \frac{e^{2}}{4 π ε_{0}}]

4

\frac{1}{C^{2}} G \frac{e^{2}}{4 π ε_{0}}

Explanation:

A Let,
$L \propto C^{x} G^{y} {(\frac{e^{2}}{4 π ε_{0}})}^{z}$
So, $[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Dimension of $[L] = [M^{0} L^{1} T^{0}]$
Dimension of $[C] = [M^{0} L^{1} T^{- 1}]$
Dimension of $[G] = [\frac{{Fr}^{2}}{m_{1} m_{2}}] = \frac{[M^{1} L^{1} T^{- 2}] [L]^{2}}{M^{2}}$
$= [M^{- 1} L^{3} T^{- 2}]$
Dimension $(\frac{e^{2}}{4 π ε_{0}}) = \frac{(I.T.)^{2}}{[M^{- 1} L^{- 3} T^{4} I^{2}]} = [M^{1} L^{3} T^{- 2} I^{0}]$
$[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Putting dimension both sides-
$[M^{0} L^{1} T^{0}] = {[M^{0} L^{1} T^{- 1}]}^{x} \cdot {[M^{- 1} L^{3} T^{- 2}]}^{y} {[M^{1} L^{3} T^{- 2}]}^{z}$
$= [M^{0} L^{x} T^{- 2 x}] [M^{- y} L^{3 y} T^{- 2 y}] [M^{z} L^{3 z} T^{- 2 z}]$
$= [M^{- y + z} L^{x + 3 y + 3 z} T^{- x - 2 y - 2 z}]$
So, $z = \frac{1}{2}$
Now, Putting the value as ' $y$ ' in eq $^{n}$ ..(iv)
$x + 6 y = 1$
$x + 6 \times \frac{1}{2} = 1$
$x + 3 = 1$
$x = - 2$
Now, Put the value of $x, y, z$
$L = {KC}^{- 2} G^{1 / 2} {(\frac{e^{2}}{4 π ε_{0}})}^{\frac{1}{2}}$
Then, $L = \frac{1}{C^{2}} {(\frac{{Ge}^{2}}{4 π ε_{0}})}^{1 / 2}$

NEET 2017

Units and Measurements

139410 In CGS system the magnitude of the force is 100 dynes. In another system, where the fundamental physical quantities are kilogram, metre and minutes. The magnitude of the force is

1 0.9

2 3.6

3 2.7

4 1.8

Explanation:

B Let, $n_{1} u_{1} = n_{2} u_{2}$
on writing their dimension both sides-
$n_{1} [M_{1} L_{1} T_{1}^{- 1}]$
$n_{2} [M_{2} L_{2} T_{2}^{- 2}]$
Now, $n_{2} = n_{1} (\frac{u_{1}}{u_{2}})$
$n_{2} = n_{1} [\frac{M_{1}}{M_{2}}] [\frac{L_{1}}{L_{2}}] [\frac{T_{1}^{- 2}}{T_{2}^{- 2}}]$
$= 100 [\frac{gm}{kg}] [\frac{cm}{m}] {[\frac{\sec}{\min}]}^{- 2}$
$- y + z = 0$
$z = y$
$x + 3 y + 3 z = 1$
$- x - 2 y - 2 z = 0$
Now putting the value of ' $z$ ' in eq ' (ii)
$x + 3 y + 3 z = 1$
$x + 3 y + 3 y = 1$
$x + 6 y = 1$
(iv)
$= 100 [\frac{1}{1000} g] [\frac{1}{100} cm] {[\frac{1}{60} \sec]}^{- 2}$
$= 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
So, $n_{2} = 3.6$

TS EAMCET 28.09.2020

Units and Measurements

139416 In a new system of units, unit of mass is $10 kg$ , unit of length is $1 km$ and unit of time is 1 minute. The value of 1 joule in this new hypothetical system is

1

3.6 \times 10^{- 4}

new units

2

6 \times 10^{7}

new units

3

10^{11}

new units

4

1.67 \times 10^{4}

new units

Explanation:

A Given, $M = 10 kg, L = 1 km, T = 1$ minute
Now, let $n_{1}$ and $n_{2}$ are the S.I unit and unit of new system.
Here, $n_{1} = 1 J$
The dimensional formula of energy is $[{ML}^{2} T^{- 2}]$
So, S.I unit of energy = new system
$n_{1} ([M_{1} L_{1}^{2} T_{1}^{- 2}]) = n_{2} ([M_{2} L_{2}^{2} T_{2}^{- 2}])$
$\frac{n_{2}}{n_{1}} = (\frac{M_{1}}{M_{2}}) {(\frac{L_{1}}{L_{2}})}^{2} (\frac{T_{1}^{- 2}}{T_{2}^{- 2}})$
$\frac{n_{2}}{n_{1}} = (\frac{1}{10}) \times {(\frac{1}{1000})}^{2} \times {(\frac{1}{60})}^{- 2}$
$\frac{n_{2}}{n_{1}} = 1 \times \frac{36}{100000}$
$n_{2} = 1 \times 3.6 \times 10^{- 4} J$
$n_{2} = 3.6 \times 10^{- 4} J$

COMEDK 2016

Units and Measurements

139408 It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun. This is called solar constant, the value of solar constant in S.I. units is

1

2 J m^{2} s^{- 1}

2

1.4 {Wm}^{- 2}

3

2.4 {kWm}^{- 2}

4

1.4 {kJm}^{- 2} s^{- 1}

Explanation:

D Given that,
It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun.
$S = 2 cal / {cm}^{2} - \min$
We know that,
$1 cal = 4.18 J$
$1 cm = 10^{- 2} m$
$1 \min = 60 \sec$
$∴ S = \frac{2 \times 4.18}{{(10^{- 2})}^{2} \times 60} \frac{J}{m^{2} - \sec}$
$S = \frac{2 \times 4.18}{10^{- 4} \times 60}$
$S = \frac{4.18 \times 10^{4}}{30}$
$S = 0.13933 \times 10^{4}$
$S = 0.14 \times 10^{4}$
$S = 1.4 \times 10^{3} \frac{J}{m^{2} - \sec}$
$S = 1.4 \frac{kJ}{m^{2} - \sec}$
Putting the value of $z$ in ${eq}^{n}$ .
$- x - 2 y - 2 z = 0$
$- x - 2 y - 2 y = 0$
$- x - 4 y = 0$
$x = - 4 y$
Now putting the value of $x$ in $e^{n}$
$x + 6 y = 1$
$- 4 y + 6 y = 1$
$2 y = 1$
$y = \frac{1}{2}$
from eq $^{n}$ (i)
$z = y$

AP EAMCET-25.04.2017

Units and Measurements

139409 A physical quantity of the dimensions of length that can be formed out of $C, G$ and $\frac{e^{2}}{4 π ε_{0}}$ is [c is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]

1

\frac{1}{C^{2}} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

2

C^{2} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

3

\frac{1}{C^{2}} [G \frac{e^{2}}{4 π ε_{0}}]

4

\frac{1}{C^{2}} G \frac{e^{2}}{4 π ε_{0}}

Explanation:

A Let,
$L \propto C^{x} G^{y} {(\frac{e^{2}}{4 π ε_{0}})}^{z}$
So, $[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Dimension of $[L] = [M^{0} L^{1} T^{0}]$
Dimension of $[C] = [M^{0} L^{1} T^{- 1}]$
Dimension of $[G] = [\frac{{Fr}^{2}}{m_{1} m_{2}}] = \frac{[M^{1} L^{1} T^{- 2}] [L]^{2}}{M^{2}}$
$= [M^{- 1} L^{3} T^{- 2}]$
Dimension $(\frac{e^{2}}{4 π ε_{0}}) = \frac{(I.T.)^{2}}{[M^{- 1} L^{- 3} T^{4} I^{2}]} = [M^{1} L^{3} T^{- 2} I^{0}]$
$[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Putting dimension both sides-
$[M^{0} L^{1} T^{0}] = {[M^{0} L^{1} T^{- 1}]}^{x} \cdot {[M^{- 1} L^{3} T^{- 2}]}^{y} {[M^{1} L^{3} T^{- 2}]}^{z}$
$= [M^{0} L^{x} T^{- 2 x}] [M^{- y} L^{3 y} T^{- 2 y}] [M^{z} L^{3 z} T^{- 2 z}]$
$= [M^{- y + z} L^{x + 3 y + 3 z} T^{- x - 2 y - 2 z}]$
So, $z = \frac{1}{2}$
Now, Putting the value as ' $y$ ' in eq $^{n}$ ..(iv)
$x + 6 y = 1$
$x + 6 \times \frac{1}{2} = 1$
$x + 3 = 1$
$x = - 2$
Now, Put the value of $x, y, z$
$L = {KC}^{- 2} G^{1 / 2} {(\frac{e^{2}}{4 π ε_{0}})}^{\frac{1}{2}}$
Then, $L = \frac{1}{C^{2}} {(\frac{{Ge}^{2}}{4 π ε_{0}})}^{1 / 2}$

NEET 2017

Units and Measurements

139410 In CGS system the magnitude of the force is 100 dynes. In another system, where the fundamental physical quantities are kilogram, metre and minutes. The magnitude of the force is

1 0.9

2 3.6

3 2.7

4 1.8

Explanation:

B Let, $n_{1} u_{1} = n_{2} u_{2}$
on writing their dimension both sides-
$n_{1} [M_{1} L_{1} T_{1}^{- 1}]$
$n_{2} [M_{2} L_{2} T_{2}^{- 2}]$
Now, $n_{2} = n_{1} (\frac{u_{1}}{u_{2}})$
$n_{2} = n_{1} [\frac{M_{1}}{M_{2}}] [\frac{L_{1}}{L_{2}}] [\frac{T_{1}^{- 2}}{T_{2}^{- 2}}]$
$= 100 [\frac{gm}{kg}] [\frac{cm}{m}] {[\frac{\sec}{\min}]}^{- 2}$
$- y + z = 0$
$z = y$
$x + 3 y + 3 z = 1$
$- x - 2 y - 2 z = 0$
Now putting the value of ' $z$ ' in eq ' (ii)
$x + 3 y + 3 z = 1$
$x + 3 y + 3 y = 1$
$x + 6 y = 1$
(iv)
$= 100 [\frac{1}{1000} g] [\frac{1}{100} cm] {[\frac{1}{60} \sec]}^{- 2}$
$= 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
So, $n_{2} = 3.6$

TS EAMCET 28.09.2020

Units and Measurements

139416 In a new system of units, unit of mass is $10 kg$ , unit of length is $1 km$ and unit of time is 1 minute. The value of 1 joule in this new hypothetical system is

1

3.6 \times 10^{- 4}

new units

2

6 \times 10^{7}

new units

3

10^{11}

new units

4

1.67 \times 10^{4}

new units

Explanation:

A Given, $M = 10 kg, L = 1 km, T = 1$ minute
Now, let $n_{1}$ and $n_{2}$ are the S.I unit and unit of new system.
Here, $n_{1} = 1 J$
The dimensional formula of energy is $[{ML}^{2} T^{- 2}]$
So, S.I unit of energy = new system
$n_{1} ([M_{1} L_{1}^{2} T_{1}^{- 2}]) = n_{2} ([M_{2} L_{2}^{2} T_{2}^{- 2}])$
$\frac{n_{2}}{n_{1}} = (\frac{M_{1}}{M_{2}}) {(\frac{L_{1}}{L_{2}})}^{2} (\frac{T_{1}^{- 2}}{T_{2}^{- 2}})$
$\frac{n_{2}}{n_{1}} = (\frac{1}{10}) \times {(\frac{1}{1000})}^{2} \times {(\frac{1}{60})}^{- 2}$
$\frac{n_{2}}{n_{1}} = 1 \times \frac{36}{100000}$
$n_{2} = 1 \times 3.6 \times 10^{- 4} J$
$n_{2} = 3.6 \times 10^{- 4} J$

COMEDK 2016

Units and Measurements

139408 It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun. This is called solar constant, the value of solar constant in S.I. units is

1

2 J m^{2} s^{- 1}

2

1.4 {Wm}^{- 2}

3

2.4 {kWm}^{- 2}

4

1.4 {kJm}^{- 2} s^{- 1}

Explanation:

D Given that,
It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun.
$S = 2 cal / {cm}^{2} - \min$
We know that,
$1 cal = 4.18 J$
$1 cm = 10^{- 2} m$
$1 \min = 60 \sec$
$∴ S = \frac{2 \times 4.18}{{(10^{- 2})}^{2} \times 60} \frac{J}{m^{2} - \sec}$
$S = \frac{2 \times 4.18}{10^{- 4} \times 60}$
$S = \frac{4.18 \times 10^{4}}{30}$
$S = 0.13933 \times 10^{4}$
$S = 0.14 \times 10^{4}$
$S = 1.4 \times 10^{3} \frac{J}{m^{2} - \sec}$
$S = 1.4 \frac{kJ}{m^{2} - \sec}$
Putting the value of $z$ in ${eq}^{n}$ .
$- x - 2 y - 2 z = 0$
$- x - 2 y - 2 y = 0$
$- x - 4 y = 0$
$x = - 4 y$
Now putting the value of $x$ in $e^{n}$
$x + 6 y = 1$
$- 4 y + 6 y = 1$
$2 y = 1$
$y = \frac{1}{2}$
from eq $^{n}$ (i)
$z = y$

AP EAMCET-25.04.2017

Units and Measurements

139409 A physical quantity of the dimensions of length that can be formed out of $C, G$ and $\frac{e^{2}}{4 π ε_{0}}$ is [c is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]

1

\frac{1}{C^{2}} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

2

C^{2} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

3

\frac{1}{C^{2}} [G \frac{e^{2}}{4 π ε_{0}}]

4

\frac{1}{C^{2}} G \frac{e^{2}}{4 π ε_{0}}

Explanation:

A Let,
$L \propto C^{x} G^{y} {(\frac{e^{2}}{4 π ε_{0}})}^{z}$
So, $[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Dimension of $[L] = [M^{0} L^{1} T^{0}]$
Dimension of $[C] = [M^{0} L^{1} T^{- 1}]$
Dimension of $[G] = [\frac{{Fr}^{2}}{m_{1} m_{2}}] = \frac{[M^{1} L^{1} T^{- 2}] [L]^{2}}{M^{2}}$
$= [M^{- 1} L^{3} T^{- 2}]$
Dimension $(\frac{e^{2}}{4 π ε_{0}}) = \frac{(I.T.)^{2}}{[M^{- 1} L^{- 3} T^{4} I^{2}]} = [M^{1} L^{3} T^{- 2} I^{0}]$
$[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Putting dimension both sides-
$[M^{0} L^{1} T^{0}] = {[M^{0} L^{1} T^{- 1}]}^{x} \cdot {[M^{- 1} L^{3} T^{- 2}]}^{y} {[M^{1} L^{3} T^{- 2}]}^{z}$
$= [M^{0} L^{x} T^{- 2 x}] [M^{- y} L^{3 y} T^{- 2 y}] [M^{z} L^{3 z} T^{- 2 z}]$
$= [M^{- y + z} L^{x + 3 y + 3 z} T^{- x - 2 y - 2 z}]$
So, $z = \frac{1}{2}$
Now, Putting the value as ' $y$ ' in eq $^{n}$ ..(iv)
$x + 6 y = 1$
$x + 6 \times \frac{1}{2} = 1$
$x + 3 = 1$
$x = - 2$
Now, Put the value of $x, y, z$
$L = {KC}^{- 2} G^{1 / 2} {(\frac{e^{2}}{4 π ε_{0}})}^{\frac{1}{2}}$
Then, $L = \frac{1}{C^{2}} {(\frac{{Ge}^{2}}{4 π ε_{0}})}^{1 / 2}$

NEET 2017

Units and Measurements

139410 In CGS system the magnitude of the force is 100 dynes. In another system, where the fundamental physical quantities are kilogram, metre and minutes. The magnitude of the force is

1 0.9

2 3.6

3 2.7

4 1.8

Explanation:

B Let, $n_{1} u_{1} = n_{2} u_{2}$
on writing their dimension both sides-
$n_{1} [M_{1} L_{1} T_{1}^{- 1}]$
$n_{2} [M_{2} L_{2} T_{2}^{- 2}]$
Now, $n_{2} = n_{1} (\frac{u_{1}}{u_{2}})$
$n_{2} = n_{1} [\frac{M_{1}}{M_{2}}] [\frac{L_{1}}{L_{2}}] [\frac{T_{1}^{- 2}}{T_{2}^{- 2}}]$
$= 100 [\frac{gm}{kg}] [\frac{cm}{m}] {[\frac{\sec}{\min}]}^{- 2}$
$- y + z = 0$
$z = y$
$x + 3 y + 3 z = 1$
$- x - 2 y - 2 z = 0$
Now putting the value of ' $z$ ' in eq ' (ii)
$x + 3 y + 3 z = 1$
$x + 3 y + 3 y = 1$
$x + 6 y = 1$
(iv)
$= 100 [\frac{1}{1000} g] [\frac{1}{100} cm] {[\frac{1}{60} \sec]}^{- 2}$
$= 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
So, $n_{2} = 3.6$

TS EAMCET 28.09.2020

Units and Measurements

139416 In a new system of units, unit of mass is $10 kg$ , unit of length is $1 km$ and unit of time is 1 minute. The value of 1 joule in this new hypothetical system is

1

3.6 \times 10^{- 4}

new units

2

6 \times 10^{7}

new units

3

10^{11}

new units

4

1.67 \times 10^{4}

new units

Explanation:

A Given, $M = 10 kg, L = 1 km, T = 1$ minute
Now, let $n_{1}$ and $n_{2}$ are the S.I unit and unit of new system.
Here, $n_{1} = 1 J$
The dimensional formula of energy is $[{ML}^{2} T^{- 2}]$
So, S.I unit of energy = new system
$n_{1} ([M_{1} L_{1}^{2} T_{1}^{- 2}]) = n_{2} ([M_{2} L_{2}^{2} T_{2}^{- 2}])$
$\frac{n_{2}}{n_{1}} = (\frac{M_{1}}{M_{2}}) {(\frac{L_{1}}{L_{2}})}^{2} (\frac{T_{1}^{- 2}}{T_{2}^{- 2}})$
$\frac{n_{2}}{n_{1}} = (\frac{1}{10}) \times {(\frac{1}{1000})}^{2} \times {(\frac{1}{60})}^{- 2}$
$\frac{n_{2}}{n_{1}} = 1 \times \frac{36}{100000}$
$n_{2} = 1 \times 3.6 \times 10^{- 4} J$
$n_{2} = 3.6 \times 10^{- 4} J$

COMEDK 2016

Units and Measurements

139408 It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun. This is called solar constant, the value of solar constant in S.I. units is

1

2 J m^{2} s^{- 1}

2

1.4 {Wm}^{- 2}

3

2.4 {kWm}^{- 2}

4

1.4 {kJm}^{- 2} s^{- 1}

Explanation:

D Given that,
It is estimated that ${cm}^{2}$ of earth receives about 2 calorie of heat energy per minute from the sun.
$S = 2 cal / {cm}^{2} - \min$
We know that,
$1 cal = 4.18 J$
$1 cm = 10^{- 2} m$
$1 \min = 60 \sec$
$∴ S = \frac{2 \times 4.18}{{(10^{- 2})}^{2} \times 60} \frac{J}{m^{2} - \sec}$
$S = \frac{2 \times 4.18}{10^{- 4} \times 60}$
$S = \frac{4.18 \times 10^{4}}{30}$
$S = 0.13933 \times 10^{4}$
$S = 0.14 \times 10^{4}$
$S = 1.4 \times 10^{3} \frac{J}{m^{2} - \sec}$
$S = 1.4 \frac{kJ}{m^{2} - \sec}$
Putting the value of $z$ in ${eq}^{n}$ .
$- x - 2 y - 2 z = 0$
$- x - 2 y - 2 y = 0$
$- x - 4 y = 0$
$x = - 4 y$
Now putting the value of $x$ in $e^{n}$
$x + 6 y = 1$
$- 4 y + 6 y = 1$
$2 y = 1$
$y = \frac{1}{2}$
from eq $^{n}$ (i)
$z = y$

AP EAMCET-25.04.2017

Units and Measurements

139409 A physical quantity of the dimensions of length that can be formed out of $C, G$ and $\frac{e^{2}}{4 π ε_{0}}$ is [c is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]

1

\frac{1}{C^{2}} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

2

C^{2} {[G \frac{e^{2}}{4 π ε_{0}}]}^{1 / 2}

3

\frac{1}{C^{2}} [G \frac{e^{2}}{4 π ε_{0}}]

4

\frac{1}{C^{2}} G \frac{e^{2}}{4 π ε_{0}}

Explanation:

A Let,
$L \propto C^{x} G^{y} {(\frac{e^{2}}{4 π ε_{0}})}^{z}$
So, $[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Dimension of $[L] = [M^{0} L^{1} T^{0}]$
Dimension of $[C] = [M^{0} L^{1} T^{- 1}]$
Dimension of $[G] = [\frac{{Fr}^{2}}{m_{1} m_{2}}] = \frac{[M^{1} L^{1} T^{- 2}] [L]^{2}}{M^{2}}$
$= [M^{- 1} L^{3} T^{- 2}]$
Dimension $(\frac{e^{2}}{4 π ε_{0}}) = \frac{(I.T.)^{2}}{[M^{- 1} L^{- 3} T^{4} I^{2}]} = [M^{1} L^{3} T^{- 2} I^{0}]$
$[L] = [C]^{x} [G]^{y} {[\frac{e^{2}}{4 π ε_{0}}]}^{z}$
Putting dimension both sides-
$[M^{0} L^{1} T^{0}] = {[M^{0} L^{1} T^{- 1}]}^{x} \cdot {[M^{- 1} L^{3} T^{- 2}]}^{y} {[M^{1} L^{3} T^{- 2}]}^{z}$
$= [M^{0} L^{x} T^{- 2 x}] [M^{- y} L^{3 y} T^{- 2 y}] [M^{z} L^{3 z} T^{- 2 z}]$
$= [M^{- y + z} L^{x + 3 y + 3 z} T^{- x - 2 y - 2 z}]$
So, $z = \frac{1}{2}$
Now, Putting the value as ' $y$ ' in eq $^{n}$ ..(iv)
$x + 6 y = 1$
$x + 6 \times \frac{1}{2} = 1$
$x + 3 = 1$
$x = - 2$
Now, Put the value of $x, y, z$
$L = {KC}^{- 2} G^{1 / 2} {(\frac{e^{2}}{4 π ε_{0}})}^{\frac{1}{2}}$
Then, $L = \frac{1}{C^{2}} {(\frac{{Ge}^{2}}{4 π ε_{0}})}^{1 / 2}$

NEET 2017

Units and Measurements

139410 In CGS system the magnitude of the force is 100 dynes. In another system, where the fundamental physical quantities are kilogram, metre and minutes. The magnitude of the force is

1 0.9

2 3.6

3 2.7

4 1.8

Explanation:

B Let, $n_{1} u_{1} = n_{2} u_{2}$
on writing their dimension both sides-
$n_{1} [M_{1} L_{1} T_{1}^{- 1}]$
$n_{2} [M_{2} L_{2} T_{2}^{- 2}]$
Now, $n_{2} = n_{1} (\frac{u_{1}}{u_{2}})$
$n_{2} = n_{1} [\frac{M_{1}}{M_{2}}] [\frac{L_{1}}{L_{2}}] [\frac{T_{1}^{- 2}}{T_{2}^{- 2}}]$
$= 100 [\frac{gm}{kg}] [\frac{cm}{m}] {[\frac{\sec}{\min}]}^{- 2}$
$- y + z = 0$
$z = y$
$x + 3 y + 3 z = 1$
$- x - 2 y - 2 z = 0$
Now putting the value of ' $z$ ' in eq ' (ii)
$x + 3 y + 3 z = 1$
$x + 3 y + 3 y = 1$
$x + 6 y = 1$
(iv)
$= 100 [\frac{1}{1000} g] [\frac{1}{100} cm] {[\frac{1}{60} \sec]}^{- 2}$
$= 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
So, $n_{2} = 3.6$

TS EAMCET 28.09.2020

Units and Measurements

139416 In a new system of units, unit of mass is $10 kg$ , unit of length is $1 km$ and unit of time is 1 minute. The value of 1 joule in this new hypothetical system is

1

3.6 \times 10^{- 4}

new units

2

6 \times 10^{7}

new units

3

10^{11}

new units

4

1.67 \times 10^{4}

new units

Explanation:

A Given, $M = 10 kg, L = 1 km, T = 1$ minute
Now, let $n_{1}$ and $n_{2}$ are the S.I unit and unit of new system.
Here, $n_{1} = 1 J$
The dimensional formula of energy is $[{ML}^{2} T^{- 2}]$
So, S.I unit of energy = new system
$n_{1} ([M_{1} L_{1}^{2} T_{1}^{- 2}]) = n_{2} ([M_{2} L_{2}^{2} T_{2}^{- 2}])$
$\frac{n_{2}}{n_{1}} = (\frac{M_{1}}{M_{2}}) {(\frac{L_{1}}{L_{2}})}^{2} (\frac{T_{1}^{- 2}}{T_{2}^{- 2}})$
$\frac{n_{2}}{n_{1}} = (\frac{1}{10}) \times {(\frac{1}{1000})}^{2} \times {(\frac{1}{60})}^{- 2}$
$\frac{n_{2}}{n_{1}} = 1 \times \frac{36}{100000}$
$n_{2} = 1 \times 3.6 \times 10^{- 4} J$
$n_{2} = 3.6 \times 10^{- 4} J$

COMEDK 2016

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