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.
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.
