NEET Test Series from KOTA - 10 Papers In MS WORD
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Units and Measurements
139500
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be
D From the relation of ohm's law- \(\mathrm{V}=\mathrm{IR}\) Where \(-\mathrm{V}=\) Voltage \(I=\) current of the circuit \(\mathrm{R}=\) Resistance of the wire or \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\) \(\therefore \mathrm{R}=\frac{\mathrm{W}}{\mathrm{q} \cdot \mathrm{I}} \left(\because \mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}\right)\) \(\text { or } (\because \mathrm{q}=\mathrm{I} \cdot \mathrm{t})\) \(\mathrm{R}=\frac{\mathrm{W}}{\mathrm{I} \cdot \mathrm{t} \cdot \mathrm{I}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{IT}][\mathrm{I}]}\) or \(\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]\)
JCECE-2018
Units and Measurements
139501
If \(M, L, T\), and \(I\) stand for mass, length, time and electric current respectively, the dimensional formula for capacitance is
B Capacitance \(=\frac{\text { charge }}{\text { voltage }}\) Capacitance \(=\) charge \(\times\) voltage \(^{-1}\) Dim. of charge \(=[\mathrm{IT}]\) Voltage \(=\) Electric field \(\times\) displacement Electric field \(=\) force \(\times\) charge \(^{-1}\) \(=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times[\mathrm{IT}]^{-1}\) Dim. \((\) Electric field \()=\left[\mathrm{ML}^{1} \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\) Now, dimension of voltage \(=\left[\mathrm{MLI}^{-1} \mathrm{~T}^{-3}\right][\mathrm{L}]\) \(=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]\) Put in the equation (i), Dim. \((\) Capacitance \()=[\mathrm{IT}] \times\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]^{-1}\) \(=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{I}^{2}\right]\)
AP EAMCET(Medical)-1997]
Units and Measurements
139506
Planck's constant has the dimensions of
1 linear momentum
2 angular momentum
3 energy
4 power
Explanation:
B We know that, \(\mathrm{E}=\mathrm{h} \nu\) Where, \(\mathrm{E}=\) Energy, \(\mathrm{h}=\) Planck's constant, \(v=\text { Frequency }\) Dimension of Plank's constant \(\mathrm{h} =\frac{\mathrm{E}}{\mathrm{v}}\) \(\mathrm{h} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]}\) \(\mathrm{h} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Dimension of Angular momentum \(=\mathrm{MVr}\) \(=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Hence, option (b) is correct.
D Magnetic flux \(\left(\phi_{\mathbf{B}}\right)=\mathrm{B} \times \mathrm{A} \times \cos \theta\) Where, \(\mathrm{B}=\) Magnetic Field \(\mathrm{A}=\) Surface Area \(\theta=\) Angle between the magnetic field and normal to the surface. Therefore, dim. \(\left(\phi_{\mathbf{B}}\right)=\left[\mathrm{M}^{1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\) \(\phi_{\mathbf{B}}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\) Since, \(\theta\) is a dimensionless quantity.
139500
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be
D From the relation of ohm's law- \(\mathrm{V}=\mathrm{IR}\) Where \(-\mathrm{V}=\) Voltage \(I=\) current of the circuit \(\mathrm{R}=\) Resistance of the wire or \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\) \(\therefore \mathrm{R}=\frac{\mathrm{W}}{\mathrm{q} \cdot \mathrm{I}} \left(\because \mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}\right)\) \(\text { or } (\because \mathrm{q}=\mathrm{I} \cdot \mathrm{t})\) \(\mathrm{R}=\frac{\mathrm{W}}{\mathrm{I} \cdot \mathrm{t} \cdot \mathrm{I}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{IT}][\mathrm{I}]}\) or \(\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]\)
JCECE-2018
Units and Measurements
139501
If \(M, L, T\), and \(I\) stand for mass, length, time and electric current respectively, the dimensional formula for capacitance is
B Capacitance \(=\frac{\text { charge }}{\text { voltage }}\) Capacitance \(=\) charge \(\times\) voltage \(^{-1}\) Dim. of charge \(=[\mathrm{IT}]\) Voltage \(=\) Electric field \(\times\) displacement Electric field \(=\) force \(\times\) charge \(^{-1}\) \(=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times[\mathrm{IT}]^{-1}\) Dim. \((\) Electric field \()=\left[\mathrm{ML}^{1} \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\) Now, dimension of voltage \(=\left[\mathrm{MLI}^{-1} \mathrm{~T}^{-3}\right][\mathrm{L}]\) \(=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]\) Put in the equation (i), Dim. \((\) Capacitance \()=[\mathrm{IT}] \times\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]^{-1}\) \(=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{I}^{2}\right]\)
AP EAMCET(Medical)-1997]
Units and Measurements
139506
Planck's constant has the dimensions of
1 linear momentum
2 angular momentum
3 energy
4 power
Explanation:
B We know that, \(\mathrm{E}=\mathrm{h} \nu\) Where, \(\mathrm{E}=\) Energy, \(\mathrm{h}=\) Planck's constant, \(v=\text { Frequency }\) Dimension of Plank's constant \(\mathrm{h} =\frac{\mathrm{E}}{\mathrm{v}}\) \(\mathrm{h} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]}\) \(\mathrm{h} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Dimension of Angular momentum \(=\mathrm{MVr}\) \(=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Hence, option (b) is correct.
D Magnetic flux \(\left(\phi_{\mathbf{B}}\right)=\mathrm{B} \times \mathrm{A} \times \cos \theta\) Where, \(\mathrm{B}=\) Magnetic Field \(\mathrm{A}=\) Surface Area \(\theta=\) Angle between the magnetic field and normal to the surface. Therefore, dim. \(\left(\phi_{\mathbf{B}}\right)=\left[\mathrm{M}^{1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\) \(\phi_{\mathbf{B}}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\) Since, \(\theta\) is a dimensionless quantity.
139500
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be
D From the relation of ohm's law- \(\mathrm{V}=\mathrm{IR}\) Where \(-\mathrm{V}=\) Voltage \(I=\) current of the circuit \(\mathrm{R}=\) Resistance of the wire or \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\) \(\therefore \mathrm{R}=\frac{\mathrm{W}}{\mathrm{q} \cdot \mathrm{I}} \left(\because \mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}\right)\) \(\text { or } (\because \mathrm{q}=\mathrm{I} \cdot \mathrm{t})\) \(\mathrm{R}=\frac{\mathrm{W}}{\mathrm{I} \cdot \mathrm{t} \cdot \mathrm{I}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{IT}][\mathrm{I}]}\) or \(\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]\)
JCECE-2018
Units and Measurements
139501
If \(M, L, T\), and \(I\) stand for mass, length, time and electric current respectively, the dimensional formula for capacitance is
B Capacitance \(=\frac{\text { charge }}{\text { voltage }}\) Capacitance \(=\) charge \(\times\) voltage \(^{-1}\) Dim. of charge \(=[\mathrm{IT}]\) Voltage \(=\) Electric field \(\times\) displacement Electric field \(=\) force \(\times\) charge \(^{-1}\) \(=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times[\mathrm{IT}]^{-1}\) Dim. \((\) Electric field \()=\left[\mathrm{ML}^{1} \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\) Now, dimension of voltage \(=\left[\mathrm{MLI}^{-1} \mathrm{~T}^{-3}\right][\mathrm{L}]\) \(=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]\) Put in the equation (i), Dim. \((\) Capacitance \()=[\mathrm{IT}] \times\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]^{-1}\) \(=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{I}^{2}\right]\)
AP EAMCET(Medical)-1997]
Units and Measurements
139506
Planck's constant has the dimensions of
1 linear momentum
2 angular momentum
3 energy
4 power
Explanation:
B We know that, \(\mathrm{E}=\mathrm{h} \nu\) Where, \(\mathrm{E}=\) Energy, \(\mathrm{h}=\) Planck's constant, \(v=\text { Frequency }\) Dimension of Plank's constant \(\mathrm{h} =\frac{\mathrm{E}}{\mathrm{v}}\) \(\mathrm{h} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]}\) \(\mathrm{h} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Dimension of Angular momentum \(=\mathrm{MVr}\) \(=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Hence, option (b) is correct.
D Magnetic flux \(\left(\phi_{\mathbf{B}}\right)=\mathrm{B} \times \mathrm{A} \times \cos \theta\) Where, \(\mathrm{B}=\) Magnetic Field \(\mathrm{A}=\) Surface Area \(\theta=\) Angle between the magnetic field and normal to the surface. Therefore, dim. \(\left(\phi_{\mathbf{B}}\right)=\left[\mathrm{M}^{1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\) \(\phi_{\mathbf{B}}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\) Since, \(\theta\) is a dimensionless quantity.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Units and Measurements
139500
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be
D From the relation of ohm's law- \(\mathrm{V}=\mathrm{IR}\) Where \(-\mathrm{V}=\) Voltage \(I=\) current of the circuit \(\mathrm{R}=\) Resistance of the wire or \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\) \(\therefore \mathrm{R}=\frac{\mathrm{W}}{\mathrm{q} \cdot \mathrm{I}} \left(\because \mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}\right)\) \(\text { or } (\because \mathrm{q}=\mathrm{I} \cdot \mathrm{t})\) \(\mathrm{R}=\frac{\mathrm{W}}{\mathrm{I} \cdot \mathrm{t} \cdot \mathrm{I}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{IT}][\mathrm{I}]}\) or \(\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]\)
JCECE-2018
Units and Measurements
139501
If \(M, L, T\), and \(I\) stand for mass, length, time and electric current respectively, the dimensional formula for capacitance is
B Capacitance \(=\frac{\text { charge }}{\text { voltage }}\) Capacitance \(=\) charge \(\times\) voltage \(^{-1}\) Dim. of charge \(=[\mathrm{IT}]\) Voltage \(=\) Electric field \(\times\) displacement Electric field \(=\) force \(\times\) charge \(^{-1}\) \(=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times[\mathrm{IT}]^{-1}\) Dim. \((\) Electric field \()=\left[\mathrm{ML}^{1} \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\) Now, dimension of voltage \(=\left[\mathrm{MLI}^{-1} \mathrm{~T}^{-3}\right][\mathrm{L}]\) \(=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]\) Put in the equation (i), Dim. \((\) Capacitance \()=[\mathrm{IT}] \times\left[\mathrm{M} \mathrm{L}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right]^{-1}\) \(=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{I}^{2}\right]\)
AP EAMCET(Medical)-1997]
Units and Measurements
139506
Planck's constant has the dimensions of
1 linear momentum
2 angular momentum
3 energy
4 power
Explanation:
B We know that, \(\mathrm{E}=\mathrm{h} \nu\) Where, \(\mathrm{E}=\) Energy, \(\mathrm{h}=\) Planck's constant, \(v=\text { Frequency }\) Dimension of Plank's constant \(\mathrm{h} =\frac{\mathrm{E}}{\mathrm{v}}\) \(\mathrm{h} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]}\) \(\mathrm{h} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Dimension of Angular momentum \(=\mathrm{MVr}\) \(=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\) Hence, option (b) is correct.
D Magnetic flux \(\left(\phi_{\mathbf{B}}\right)=\mathrm{B} \times \mathrm{A} \times \cos \theta\) Where, \(\mathrm{B}=\) Magnetic Field \(\mathrm{A}=\) Surface Area \(\theta=\) Angle between the magnetic field and normal to the surface. Therefore, dim. \(\left(\phi_{\mathbf{B}}\right)=\left[\mathrm{M}^{1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\) \(\phi_{\mathbf{B}}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\) Since, \(\theta\) is a dimensionless quantity.