139419
The following physical quantities has a ratio of \(10^{3}\) between its SI units and CGS units
1 Universal gravitational constant
2 Boltzman's constant
3 Planck's constant
4 Young's modulus of elasticity
Explanation:
A Universal gravitational constant \(\text { SI unit }=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}}\) \(\text { CGS unit }=\frac{\text { dyne- }-\mathrm{cm}^{2}}{\mathrm{gm}^{2}}\) \(\text { Ratio }=\frac{\text { SI unit }}{\text { CGSunit }}=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\mathrm{gm}^{2}}{\text { dyne }-\mathrm{cm}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\left(10^{-3} \mathrm{~kg}\right)^{2}}{10^{-5} \mathrm{~N} \times 10^{-4} \mathrm{~m}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{10^{-6} \mathrm{~kg}^{2}}{10^{-9} \mathrm{~N}-\mathrm{m}^{2}}\) \(=10^{-6+9}\) \(=10^{3}\) Hence, the ratio of universal gravitational constant has \(10^{3}\) between its SI unit and CGS units.
EAMCET-1994
Units and Measurements
139420
In a particular system, the unit of length, mass and time are chosen to be \(10 \mathrm{~cm}, 10 \mathrm{~g}\) and \(0.1 \mathrm{~s}\) respectively. The unit of force in this system will be equivalent to
1 \(0.1 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(100 \mathrm{~N}\)
Explanation:
A We know that \(1 \mathrm{~N}=\frac{1 \mathrm{~kg} \times 1 \mathrm{~m}}{1-\mathrm{s}^{2}}\) \(=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1-\mathrm{s}^{2}}\) \(=\frac{100 \times(10 \mathrm{~g}) \times 10 \times(10 \mathrm{~cm})}{100 \times(0.1)^{2}}\) \(\text { So, } 1 \mathrm{~N}=10 \mathrm{~F}_{\text {new }}\left(\mathrm{F}_{\text {new }}\right. \text { is new unit of force) }\) \(\mathrm{F}_{\text {new }}=\frac{1}{10} \mathrm{~N}=0.1 \mathrm{~N}\)
AIPMT 1994
Units and Measurements
139422
In a new system of units called star units, \(1 \mathrm{~kg}^{*}=10 \mathrm{~kg}, 1 \mathrm{~m}^{*}=1 \mathrm{~km}\) and \(1 \mathrm{~s}^{*}=1\) minute, what will be the value of \(1 \mathrm{~J}\) in the new system
139423
A new unit of length is so chosen that the speed of light in vacuum is unity. Calculate the distance (in this new unit) between the sun and the earth if light takes \(8 \mathrm{~min}\) and 20 seconds to reach earth from sun.
1 300
2 400
3 500
4 600
Explanation:
C Given that, The speed of light in a vacuum is unity, \(\mathrm{v}=1 \mathrm{~m} / \mathrm{sec}\) Time \(=8 \mathrm{~min}\) and \(20 \mathrm{sec}=500 \mathrm{sec}\) We know that, Distance \(=\) speed \(\times\) time Distance \(=1 \times 500\) Distance \(=500\) unit Hence, distance between sun and earth is 500 unit.
139419
The following physical quantities has a ratio of \(10^{3}\) between its SI units and CGS units
1 Universal gravitational constant
2 Boltzman's constant
3 Planck's constant
4 Young's modulus of elasticity
Explanation:
A Universal gravitational constant \(\text { SI unit }=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}}\) \(\text { CGS unit }=\frac{\text { dyne- }-\mathrm{cm}^{2}}{\mathrm{gm}^{2}}\) \(\text { Ratio }=\frac{\text { SI unit }}{\text { CGSunit }}=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\mathrm{gm}^{2}}{\text { dyne }-\mathrm{cm}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\left(10^{-3} \mathrm{~kg}\right)^{2}}{10^{-5} \mathrm{~N} \times 10^{-4} \mathrm{~m}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{10^{-6} \mathrm{~kg}^{2}}{10^{-9} \mathrm{~N}-\mathrm{m}^{2}}\) \(=10^{-6+9}\) \(=10^{3}\) Hence, the ratio of universal gravitational constant has \(10^{3}\) between its SI unit and CGS units.
EAMCET-1994
Units and Measurements
139420
In a particular system, the unit of length, mass and time are chosen to be \(10 \mathrm{~cm}, 10 \mathrm{~g}\) and \(0.1 \mathrm{~s}\) respectively. The unit of force in this system will be equivalent to
1 \(0.1 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(100 \mathrm{~N}\)
Explanation:
A We know that \(1 \mathrm{~N}=\frac{1 \mathrm{~kg} \times 1 \mathrm{~m}}{1-\mathrm{s}^{2}}\) \(=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1-\mathrm{s}^{2}}\) \(=\frac{100 \times(10 \mathrm{~g}) \times 10 \times(10 \mathrm{~cm})}{100 \times(0.1)^{2}}\) \(\text { So, } 1 \mathrm{~N}=10 \mathrm{~F}_{\text {new }}\left(\mathrm{F}_{\text {new }}\right. \text { is new unit of force) }\) \(\mathrm{F}_{\text {new }}=\frac{1}{10} \mathrm{~N}=0.1 \mathrm{~N}\)
AIPMT 1994
Units and Measurements
139422
In a new system of units called star units, \(1 \mathrm{~kg}^{*}=10 \mathrm{~kg}, 1 \mathrm{~m}^{*}=1 \mathrm{~km}\) and \(1 \mathrm{~s}^{*}=1\) minute, what will be the value of \(1 \mathrm{~J}\) in the new system
139423
A new unit of length is so chosen that the speed of light in vacuum is unity. Calculate the distance (in this new unit) between the sun and the earth if light takes \(8 \mathrm{~min}\) and 20 seconds to reach earth from sun.
1 300
2 400
3 500
4 600
Explanation:
C Given that, The speed of light in a vacuum is unity, \(\mathrm{v}=1 \mathrm{~m} / \mathrm{sec}\) Time \(=8 \mathrm{~min}\) and \(20 \mathrm{sec}=500 \mathrm{sec}\) We know that, Distance \(=\) speed \(\times\) time Distance \(=1 \times 500\) Distance \(=500\) unit Hence, distance between sun and earth is 500 unit.
139419
The following physical quantities has a ratio of \(10^{3}\) between its SI units and CGS units
1 Universal gravitational constant
2 Boltzman's constant
3 Planck's constant
4 Young's modulus of elasticity
Explanation:
A Universal gravitational constant \(\text { SI unit }=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}}\) \(\text { CGS unit }=\frac{\text { dyne- }-\mathrm{cm}^{2}}{\mathrm{gm}^{2}}\) \(\text { Ratio }=\frac{\text { SI unit }}{\text { CGSunit }}=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\mathrm{gm}^{2}}{\text { dyne }-\mathrm{cm}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\left(10^{-3} \mathrm{~kg}\right)^{2}}{10^{-5} \mathrm{~N} \times 10^{-4} \mathrm{~m}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{10^{-6} \mathrm{~kg}^{2}}{10^{-9} \mathrm{~N}-\mathrm{m}^{2}}\) \(=10^{-6+9}\) \(=10^{3}\) Hence, the ratio of universal gravitational constant has \(10^{3}\) between its SI unit and CGS units.
EAMCET-1994
Units and Measurements
139420
In a particular system, the unit of length, mass and time are chosen to be \(10 \mathrm{~cm}, 10 \mathrm{~g}\) and \(0.1 \mathrm{~s}\) respectively. The unit of force in this system will be equivalent to
1 \(0.1 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(100 \mathrm{~N}\)
Explanation:
A We know that \(1 \mathrm{~N}=\frac{1 \mathrm{~kg} \times 1 \mathrm{~m}}{1-\mathrm{s}^{2}}\) \(=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1-\mathrm{s}^{2}}\) \(=\frac{100 \times(10 \mathrm{~g}) \times 10 \times(10 \mathrm{~cm})}{100 \times(0.1)^{2}}\) \(\text { So, } 1 \mathrm{~N}=10 \mathrm{~F}_{\text {new }}\left(\mathrm{F}_{\text {new }}\right. \text { is new unit of force) }\) \(\mathrm{F}_{\text {new }}=\frac{1}{10} \mathrm{~N}=0.1 \mathrm{~N}\)
AIPMT 1994
Units and Measurements
139422
In a new system of units called star units, \(1 \mathrm{~kg}^{*}=10 \mathrm{~kg}, 1 \mathrm{~m}^{*}=1 \mathrm{~km}\) and \(1 \mathrm{~s}^{*}=1\) minute, what will be the value of \(1 \mathrm{~J}\) in the new system
139423
A new unit of length is so chosen that the speed of light in vacuum is unity. Calculate the distance (in this new unit) between the sun and the earth if light takes \(8 \mathrm{~min}\) and 20 seconds to reach earth from sun.
1 300
2 400
3 500
4 600
Explanation:
C Given that, The speed of light in a vacuum is unity, \(\mathrm{v}=1 \mathrm{~m} / \mathrm{sec}\) Time \(=8 \mathrm{~min}\) and \(20 \mathrm{sec}=500 \mathrm{sec}\) We know that, Distance \(=\) speed \(\times\) time Distance \(=1 \times 500\) Distance \(=500\) unit Hence, distance between sun and earth is 500 unit.
139419
The following physical quantities has a ratio of \(10^{3}\) between its SI units and CGS units
1 Universal gravitational constant
2 Boltzman's constant
3 Planck's constant
4 Young's modulus of elasticity
Explanation:
A Universal gravitational constant \(\text { SI unit }=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}}\) \(\text { CGS unit }=\frac{\text { dyne- }-\mathrm{cm}^{2}}{\mathrm{gm}^{2}}\) \(\text { Ratio }=\frac{\text { SI unit }}{\text { CGSunit }}=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\mathrm{gm}^{2}}{\text { dyne }-\mathrm{cm}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{\left(10^{-3} \mathrm{~kg}\right)^{2}}{10^{-5} \mathrm{~N} \times 10^{-4} \mathrm{~m}^{2}}\) \(=\frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{~kg}^{2}} \times \frac{10^{-6} \mathrm{~kg}^{2}}{10^{-9} \mathrm{~N}-\mathrm{m}^{2}}\) \(=10^{-6+9}\) \(=10^{3}\) Hence, the ratio of universal gravitational constant has \(10^{3}\) between its SI unit and CGS units.
EAMCET-1994
Units and Measurements
139420
In a particular system, the unit of length, mass and time are chosen to be \(10 \mathrm{~cm}, 10 \mathrm{~g}\) and \(0.1 \mathrm{~s}\) respectively. The unit of force in this system will be equivalent to
1 \(0.1 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(10 \mathrm{~N}\)
4 \(100 \mathrm{~N}\)
Explanation:
A We know that \(1 \mathrm{~N}=\frac{1 \mathrm{~kg} \times 1 \mathrm{~m}}{1-\mathrm{s}^{2}}\) \(=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1-\mathrm{s}^{2}}\) \(=\frac{100 \times(10 \mathrm{~g}) \times 10 \times(10 \mathrm{~cm})}{100 \times(0.1)^{2}}\) \(\text { So, } 1 \mathrm{~N}=10 \mathrm{~F}_{\text {new }}\left(\mathrm{F}_{\text {new }}\right. \text { is new unit of force) }\) \(\mathrm{F}_{\text {new }}=\frac{1}{10} \mathrm{~N}=0.1 \mathrm{~N}\)
AIPMT 1994
Units and Measurements
139422
In a new system of units called star units, \(1 \mathrm{~kg}^{*}=10 \mathrm{~kg}, 1 \mathrm{~m}^{*}=1 \mathrm{~km}\) and \(1 \mathrm{~s}^{*}=1\) minute, what will be the value of \(1 \mathrm{~J}\) in the new system
139423
A new unit of length is so chosen that the speed of light in vacuum is unity. Calculate the distance (in this new unit) between the sun and the earth if light takes \(8 \mathrm{~min}\) and 20 seconds to reach earth from sun.
1 300
2 400
3 500
4 600
Explanation:
C Given that, The speed of light in a vacuum is unity, \(\mathrm{v}=1 \mathrm{~m} / \mathrm{sec}\) Time \(=8 \mathrm{~min}\) and \(20 \mathrm{sec}=500 \mathrm{sec}\) We know that, Distance \(=\) speed \(\times\) time Distance \(=1 \times 500\) Distance \(=500\) unit Hence, distance between sun and earth is 500 unit.
