139399
The angle of \(1^{\prime}\) (minute of arc) in radian is nearly equal to
1 \(2.91 \times 10^{-4} \mathrm{rad}\)
2 \(4.85 \times 10^{-4} \mathrm{rad}\)
3 \(4.80 \times 10^{-6} \mathrm{rad}\)
4 \(1.75 \times 10^{-2} \mathrm{rad}\)
Explanation:
A We know that, \(1^{\prime}(\) minute of arc \()=\left(\frac{1}{60}\right)\) \(=\frac{1}{60} \times \frac{\pi}{180}\) radian \(=\frac{\pi}{60 \times 180}\) \(=\frac{3.14}{10800}\) \(=0.0291 \times 10^{-2}\) \(=2.91 \times 10^{-4}\) radian
NEET (Oct.) 2020
Units and Measurements
139401
The density of a material in SI units is \(128 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). In certain units in which the unit of length is \(25 \mathrm{~cm}\) and the unit of mass is \(50 \mathrm{~g}\), the numerical value of density of the material is
1 40
2 16
3 640
4 410
Explanation:
A We know that Density \(=\frac{\text { Mass }}{\text { Volume }}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\mathrm{L}^{3}}\) Then, \(\mathrm{n}_{1} \mathrm{u}_{1}=\mathrm{n}_{2} \mathrm{u}_{2}\) \(128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]=\mathrm{n}_{2}\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]\) \(\mathrm{n}_{2}=\frac{128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]}\) \(\mathrm{n}_{1}=\) numerical value in S.I unit \(\mathrm{u}_{1}=\) unit in S.I. \(\mathrm{n}_{2}=\) numerical value in other unit \(\mathrm{u}_{2}=\) unit in other system In second unit system \(\mathrm{m}=50 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}\). \(\mathrm{L}=25 \mathrm{~cm}=\frac{25}{100} \mathrm{mtr}\) \(\mathrm{n}_{2}=128\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right]^{3}\) \(=128 \times\left[\frac{1}{\frac{50}{1000}}\right] \times\left[\frac{\frac{25}{100}}{1}\right]^{3}\) \(=128 \times 20 \times\left(\frac{1}{4}\right)^{3}\) \(=128 \times 20 \times\left(\frac{1}{64}\right)\) \(=40\)
JEE Main-10.01.2019
Units and Measurements
139404
What is dimensions of energy in terms of linear momentum (P), area of \((\mathrm{A})\) and Time (T)
C Let, \(\text { Energy } \mathrm{E}=\mathrm{kP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\) Where \(\mathrm{k}\) is a dimensionless constant of proportionalityWriting dimension on both sides- \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]^{\mathrm{c}}\) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}+\mathrm{c}}\right]\) Comparing the power from both sides- \(a=1\) \(a+2 b=2\) \(-a+c=-2\) On solving (ii), (iii) and (iv), we have- \(\begin{array}{ll} \mathrm{a}=1, \mathrm{~b}=1 / 2, \mathrm{c}= \\ \text { So, } {[\mathrm{E}]=\left[\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]}\end{array}\)
JIPMER-2019
Units and Measurements
139405
If \(10 \mathrm{~g} \mathrm{cms}^{-1}=\mathbf{x}\) 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 \(1 \mathrm{~g}=10^{-3} \mathrm{~kg}\) \(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) According to question- \(10 \mathrm{gcms}^{-1}=10^{-3} \mathrm{~kg} \times 10^{-2} \mathrm{~m} \times \mathrm{sec}^{-1}\) \(=10^{-5} \mathrm{~kg} \mathrm{~m} \cdot \mathrm{sec}^{-1}\) \(= 10^{-5} \mathrm{Ns} \quad\left[\mathrm{N}=\frac{\mathrm{kgm}}{\mathrm{sec}}\right]\)
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.
A We know that One-light year is the distance Travelled by light in one year. So, light year \(=9.46 \times 10^{12} \mathrm{~km}=9.46 \times 10^{15} \mathrm{~m}\) So, \(10 \text { light year } =9.4 \times 10^{15} \times 10 \mathrm{~m}\) \(=9.46 \times 10^{16} \mathrm{~m}\)
139399
The angle of \(1^{\prime}\) (minute of arc) in radian is nearly equal to
1 \(2.91 \times 10^{-4} \mathrm{rad}\)
2 \(4.85 \times 10^{-4} \mathrm{rad}\)
3 \(4.80 \times 10^{-6} \mathrm{rad}\)
4 \(1.75 \times 10^{-2} \mathrm{rad}\)
Explanation:
A We know that, \(1^{\prime}(\) minute of arc \()=\left(\frac{1}{60}\right)\) \(=\frac{1}{60} \times \frac{\pi}{180}\) radian \(=\frac{\pi}{60 \times 180}\) \(=\frac{3.14}{10800}\) \(=0.0291 \times 10^{-2}\) \(=2.91 \times 10^{-4}\) radian
NEET (Oct.) 2020
Units and Measurements
139401
The density of a material in SI units is \(128 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). In certain units in which the unit of length is \(25 \mathrm{~cm}\) and the unit of mass is \(50 \mathrm{~g}\), the numerical value of density of the material is
1 40
2 16
3 640
4 410
Explanation:
A We know that Density \(=\frac{\text { Mass }}{\text { Volume }}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\mathrm{L}^{3}}\) Then, \(\mathrm{n}_{1} \mathrm{u}_{1}=\mathrm{n}_{2} \mathrm{u}_{2}\) \(128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]=\mathrm{n}_{2}\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]\) \(\mathrm{n}_{2}=\frac{128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]}\) \(\mathrm{n}_{1}=\) numerical value in S.I unit \(\mathrm{u}_{1}=\) unit in S.I. \(\mathrm{n}_{2}=\) numerical value in other unit \(\mathrm{u}_{2}=\) unit in other system In second unit system \(\mathrm{m}=50 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}\). \(\mathrm{L}=25 \mathrm{~cm}=\frac{25}{100} \mathrm{mtr}\) \(\mathrm{n}_{2}=128\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right]^{3}\) \(=128 \times\left[\frac{1}{\frac{50}{1000}}\right] \times\left[\frac{\frac{25}{100}}{1}\right]^{3}\) \(=128 \times 20 \times\left(\frac{1}{4}\right)^{3}\) \(=128 \times 20 \times\left(\frac{1}{64}\right)\) \(=40\)
JEE Main-10.01.2019
Units and Measurements
139404
What is dimensions of energy in terms of linear momentum (P), area of \((\mathrm{A})\) and Time (T)
C Let, \(\text { Energy } \mathrm{E}=\mathrm{kP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\) Where \(\mathrm{k}\) is a dimensionless constant of proportionalityWriting dimension on both sides- \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]^{\mathrm{c}}\) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}+\mathrm{c}}\right]\) Comparing the power from both sides- \(a=1\) \(a+2 b=2\) \(-a+c=-2\) On solving (ii), (iii) and (iv), we have- \(\begin{array}{ll} \mathrm{a}=1, \mathrm{~b}=1 / 2, \mathrm{c}= \\ \text { So, } {[\mathrm{E}]=\left[\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]}\end{array}\)
JIPMER-2019
Units and Measurements
139405
If \(10 \mathrm{~g} \mathrm{cms}^{-1}=\mathbf{x}\) 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 \(1 \mathrm{~g}=10^{-3} \mathrm{~kg}\) \(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) According to question- \(10 \mathrm{gcms}^{-1}=10^{-3} \mathrm{~kg} \times 10^{-2} \mathrm{~m} \times \mathrm{sec}^{-1}\) \(=10^{-5} \mathrm{~kg} \mathrm{~m} \cdot \mathrm{sec}^{-1}\) \(= 10^{-5} \mathrm{Ns} \quad\left[\mathrm{N}=\frac{\mathrm{kgm}}{\mathrm{sec}}\right]\)
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.
A We know that One-light year is the distance Travelled by light in one year. So, light year \(=9.46 \times 10^{12} \mathrm{~km}=9.46 \times 10^{15} \mathrm{~m}\) So, \(10 \text { light year } =9.4 \times 10^{15} \times 10 \mathrm{~m}\) \(=9.46 \times 10^{16} \mathrm{~m}\)
139399
The angle of \(1^{\prime}\) (minute of arc) in radian is nearly equal to
1 \(2.91 \times 10^{-4} \mathrm{rad}\)
2 \(4.85 \times 10^{-4} \mathrm{rad}\)
3 \(4.80 \times 10^{-6} \mathrm{rad}\)
4 \(1.75 \times 10^{-2} \mathrm{rad}\)
Explanation:
A We know that, \(1^{\prime}(\) minute of arc \()=\left(\frac{1}{60}\right)\) \(=\frac{1}{60} \times \frac{\pi}{180}\) radian \(=\frac{\pi}{60 \times 180}\) \(=\frac{3.14}{10800}\) \(=0.0291 \times 10^{-2}\) \(=2.91 \times 10^{-4}\) radian
NEET (Oct.) 2020
Units and Measurements
139401
The density of a material in SI units is \(128 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). In certain units in which the unit of length is \(25 \mathrm{~cm}\) and the unit of mass is \(50 \mathrm{~g}\), the numerical value of density of the material is
1 40
2 16
3 640
4 410
Explanation:
A We know that Density \(=\frac{\text { Mass }}{\text { Volume }}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\mathrm{L}^{3}}\) Then, \(\mathrm{n}_{1} \mathrm{u}_{1}=\mathrm{n}_{2} \mathrm{u}_{2}\) \(128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]=\mathrm{n}_{2}\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]\) \(\mathrm{n}_{2}=\frac{128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]}\) \(\mathrm{n}_{1}=\) numerical value in S.I unit \(\mathrm{u}_{1}=\) unit in S.I. \(\mathrm{n}_{2}=\) numerical value in other unit \(\mathrm{u}_{2}=\) unit in other system In second unit system \(\mathrm{m}=50 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}\). \(\mathrm{L}=25 \mathrm{~cm}=\frac{25}{100} \mathrm{mtr}\) \(\mathrm{n}_{2}=128\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right]^{3}\) \(=128 \times\left[\frac{1}{\frac{50}{1000}}\right] \times\left[\frac{\frac{25}{100}}{1}\right]^{3}\) \(=128 \times 20 \times\left(\frac{1}{4}\right)^{3}\) \(=128 \times 20 \times\left(\frac{1}{64}\right)\) \(=40\)
JEE Main-10.01.2019
Units and Measurements
139404
What is dimensions of energy in terms of linear momentum (P), area of \((\mathrm{A})\) and Time (T)
C Let, \(\text { Energy } \mathrm{E}=\mathrm{kP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\) Where \(\mathrm{k}\) is a dimensionless constant of proportionalityWriting dimension on both sides- \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]^{\mathrm{c}}\) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}+\mathrm{c}}\right]\) Comparing the power from both sides- \(a=1\) \(a+2 b=2\) \(-a+c=-2\) On solving (ii), (iii) and (iv), we have- \(\begin{array}{ll} \mathrm{a}=1, \mathrm{~b}=1 / 2, \mathrm{c}= \\ \text { So, } {[\mathrm{E}]=\left[\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]}\end{array}\)
JIPMER-2019
Units and Measurements
139405
If \(10 \mathrm{~g} \mathrm{cms}^{-1}=\mathbf{x}\) 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 \(1 \mathrm{~g}=10^{-3} \mathrm{~kg}\) \(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) According to question- \(10 \mathrm{gcms}^{-1}=10^{-3} \mathrm{~kg} \times 10^{-2} \mathrm{~m} \times \mathrm{sec}^{-1}\) \(=10^{-5} \mathrm{~kg} \mathrm{~m} \cdot \mathrm{sec}^{-1}\) \(= 10^{-5} \mathrm{Ns} \quad\left[\mathrm{N}=\frac{\mathrm{kgm}}{\mathrm{sec}}\right]\)
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.
A We know that One-light year is the distance Travelled by light in one year. So, light year \(=9.46 \times 10^{12} \mathrm{~km}=9.46 \times 10^{15} \mathrm{~m}\) So, \(10 \text { light year } =9.4 \times 10^{15} \times 10 \mathrm{~m}\) \(=9.46 \times 10^{16} \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Units and Measurements
139399
The angle of \(1^{\prime}\) (minute of arc) in radian is nearly equal to
1 \(2.91 \times 10^{-4} \mathrm{rad}\)
2 \(4.85 \times 10^{-4} \mathrm{rad}\)
3 \(4.80 \times 10^{-6} \mathrm{rad}\)
4 \(1.75 \times 10^{-2} \mathrm{rad}\)
Explanation:
A We know that, \(1^{\prime}(\) minute of arc \()=\left(\frac{1}{60}\right)\) \(=\frac{1}{60} \times \frac{\pi}{180}\) radian \(=\frac{\pi}{60 \times 180}\) \(=\frac{3.14}{10800}\) \(=0.0291 \times 10^{-2}\) \(=2.91 \times 10^{-4}\) radian
NEET (Oct.) 2020
Units and Measurements
139401
The density of a material in SI units is \(128 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). In certain units in which the unit of length is \(25 \mathrm{~cm}\) and the unit of mass is \(50 \mathrm{~g}\), the numerical value of density of the material is
1 40
2 16
3 640
4 410
Explanation:
A We know that Density \(=\frac{\text { Mass }}{\text { Volume }}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\mathrm{L}^{3}}\) Then, \(\mathrm{n}_{1} \mathrm{u}_{1}=\mathrm{n}_{2} \mathrm{u}_{2}\) \(128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]=\mathrm{n}_{2}\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]\) \(\mathrm{n}_{2}=\frac{128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]}\) \(\mathrm{n}_{1}=\) numerical value in S.I unit \(\mathrm{u}_{1}=\) unit in S.I. \(\mathrm{n}_{2}=\) numerical value in other unit \(\mathrm{u}_{2}=\) unit in other system In second unit system \(\mathrm{m}=50 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}\). \(\mathrm{L}=25 \mathrm{~cm}=\frac{25}{100} \mathrm{mtr}\) \(\mathrm{n}_{2}=128\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right]^{3}\) \(=128 \times\left[\frac{1}{\frac{50}{1000}}\right] \times\left[\frac{\frac{25}{100}}{1}\right]^{3}\) \(=128 \times 20 \times\left(\frac{1}{4}\right)^{3}\) \(=128 \times 20 \times\left(\frac{1}{64}\right)\) \(=40\)
JEE Main-10.01.2019
Units and Measurements
139404
What is dimensions of energy in terms of linear momentum (P), area of \((\mathrm{A})\) and Time (T)
C Let, \(\text { Energy } \mathrm{E}=\mathrm{kP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\) Where \(\mathrm{k}\) is a dimensionless constant of proportionalityWriting dimension on both sides- \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]^{\mathrm{c}}\) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}+\mathrm{c}}\right]\) Comparing the power from both sides- \(a=1\) \(a+2 b=2\) \(-a+c=-2\) On solving (ii), (iii) and (iv), we have- \(\begin{array}{ll} \mathrm{a}=1, \mathrm{~b}=1 / 2, \mathrm{c}= \\ \text { So, } {[\mathrm{E}]=\left[\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]}\end{array}\)
JIPMER-2019
Units and Measurements
139405
If \(10 \mathrm{~g} \mathrm{cms}^{-1}=\mathbf{x}\) 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 \(1 \mathrm{~g}=10^{-3} \mathrm{~kg}\) \(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) According to question- \(10 \mathrm{gcms}^{-1}=10^{-3} \mathrm{~kg} \times 10^{-2} \mathrm{~m} \times \mathrm{sec}^{-1}\) \(=10^{-5} \mathrm{~kg} \mathrm{~m} \cdot \mathrm{sec}^{-1}\) \(= 10^{-5} \mathrm{Ns} \quad\left[\mathrm{N}=\frac{\mathrm{kgm}}{\mathrm{sec}}\right]\)
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.
A We know that One-light year is the distance Travelled by light in one year. So, light year \(=9.46 \times 10^{12} \mathrm{~km}=9.46 \times 10^{15} \mathrm{~m}\) So, \(10 \text { light year } =9.4 \times 10^{15} \times 10 \mathrm{~m}\) \(=9.46 \times 10^{16} \mathrm{~m}\)
139399
The angle of \(1^{\prime}\) (minute of arc) in radian is nearly equal to
1 \(2.91 \times 10^{-4} \mathrm{rad}\)
2 \(4.85 \times 10^{-4} \mathrm{rad}\)
3 \(4.80 \times 10^{-6} \mathrm{rad}\)
4 \(1.75 \times 10^{-2} \mathrm{rad}\)
Explanation:
A We know that, \(1^{\prime}(\) minute of arc \()=\left(\frac{1}{60}\right)\) \(=\frac{1}{60} \times \frac{\pi}{180}\) radian \(=\frac{\pi}{60 \times 180}\) \(=\frac{3.14}{10800}\) \(=0.0291 \times 10^{-2}\) \(=2.91 \times 10^{-4}\) radian
NEET (Oct.) 2020
Units and Measurements
139401
The density of a material in SI units is \(128 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). In certain units in which the unit of length is \(25 \mathrm{~cm}\) and the unit of mass is \(50 \mathrm{~g}\), the numerical value of density of the material is
1 40
2 16
3 640
4 410
Explanation:
A We know that Density \(=\frac{\text { Mass }}{\text { Volume }}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{M}}{\mathrm{L}^{3}}\) Then, \(\mathrm{n}_{1} \mathrm{u}_{1}=\mathrm{n}_{2} \mathrm{u}_{2}\) \(128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]=\mathrm{n}_{2}\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]\) \(\mathrm{n}_{2}=\frac{128\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{-3}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{-3}\right]}\) \(\mathrm{n}_{1}=\) numerical value in S.I unit \(\mathrm{u}_{1}=\) unit in S.I. \(\mathrm{n}_{2}=\) numerical value in other unit \(\mathrm{u}_{2}=\) unit in other system In second unit system \(\mathrm{m}=50 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}\). \(\mathrm{L}=25 \mathrm{~cm}=\frac{25}{100} \mathrm{mtr}\) \(\mathrm{n}_{2}=128\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right]^{3}\) \(=128 \times\left[\frac{1}{\frac{50}{1000}}\right] \times\left[\frac{\frac{25}{100}}{1}\right]^{3}\) \(=128 \times 20 \times\left(\frac{1}{4}\right)^{3}\) \(=128 \times 20 \times\left(\frac{1}{64}\right)\) \(=40\)
JEE Main-10.01.2019
Units and Measurements
139404
What is dimensions of energy in terms of linear momentum (P), area of \((\mathrm{A})\) and Time (T)
C Let, \(\text { Energy } \mathrm{E}=\mathrm{kP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\) Where \(\mathrm{k}\) is a dimensionless constant of proportionalityWriting dimension on both sides- \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]^{\mathrm{c}}\) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}+\mathrm{c}}\right]\) Comparing the power from both sides- \(a=1\) \(a+2 b=2\) \(-a+c=-2\) On solving (ii), (iii) and (iv), we have- \(\begin{array}{ll} \mathrm{a}=1, \mathrm{~b}=1 / 2, \mathrm{c}= \\ \text { So, } {[\mathrm{E}]=\left[\mathrm{P}^{1} \mathrm{~A}^{1 / 2} \mathrm{~T}^{-1}\right]}\end{array}\)
JIPMER-2019
Units and Measurements
139405
If \(10 \mathrm{~g} \mathrm{cms}^{-1}=\mathbf{x}\) 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 \(1 \mathrm{~g}=10^{-3} \mathrm{~kg}\) \(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) According to question- \(10 \mathrm{gcms}^{-1}=10^{-3} \mathrm{~kg} \times 10^{-2} \mathrm{~m} \times \mathrm{sec}^{-1}\) \(=10^{-5} \mathrm{~kg} \mathrm{~m} \cdot \mathrm{sec}^{-1}\) \(= 10^{-5} \mathrm{Ns} \quad\left[\mathrm{N}=\frac{\mathrm{kgm}}{\mathrm{sec}}\right]\)
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.
A We know that One-light year is the distance Travelled by light in one year. So, light year \(=9.46 \times 10^{12} \mathrm{~km}=9.46 \times 10^{15} \mathrm{~m}\) So, \(10 \text { light year } =9.4 \times 10^{15} \times 10 \mathrm{~m}\) \(=9.46 \times 10^{16} \mathrm{~m}\)