139469
The velocity of a particle is given by \(v=a t^{2}+b t\) \(+c\) If \(v\) is measured in \(\mathrm{ms}^{-1}\) and \(t\) is measured in second, the unit of
1 \(a\) is \(\mathrm{ms}^{-1}\)
2 \(b\) is \(\mathrm{ms}^{-1}\)
3 \(\mathrm{c}\) is \(\mathrm{ms}^{-1}\)
4 a and b is same but that of c is different
Explanation:
C Given, Then, \(\mathrm{v}=\mathrm{at}^{2}+\mathrm{bt}+\mathrm{c}, \mathrm{v}=\mathrm{ms}^{-1} \text { and } \mathrm{t}=\text { second }\) Unit of \(\mathrm{v}=\) Unit of \(a t^{2}\) \(V=a t^{2}\) \(\mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}^{2}}=\frac{\mathrm{ms}^{-1}}{\text { Second }^{2}}=\mathrm{ms}^{-3}\) Unit of \(v=\) Unit of \(b t\) \(v=b t\) \(\mathrm{b}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{\mathrm{ms}^{-1}}{\text { Second }}=\mathrm{ms}^{-2}\) Unit of \(\mathrm{v}=\) Unit of \(\mathrm{c}\) \(\mathrm{c}=\mathrm{v}=\mathrm{ms}^{-1}\)
CG PET- 2010
Units and Measurements
139476
SI unit of coefficient of viscosity is
1 \(\mathrm{Nsm}^{-2}\)
2 \(\mathrm{Ns}^{-1} \mathrm{~m}^{2}\)
3 \(\mathrm{Nsm}^{-3}\)
4 \(\mathrm{Ns}^{-2} \mathrm{~m}\)
Explanation:
A The Co-efficient of viscosity \(\eta\) is defined as the tangential force \(F\) required to maintain a unit velocity gradient between two parallel layer of liquid of unit area \(\mathrm{A}\). Co-efficient of viscosity \((\eta)=\frac{\mathrm{Fr}}{\mathrm{AV}}\) Where, \(\mathrm{F}=\) Force \(\mathrm{A}=\) Area \(\mathrm{V}=\) Velocity \(\mathrm{r}=\) Distance between the layers \(\eta=\frac{\mathrm{N}-\mathrm{m}}{\mathrm{m}^{2} \times \mathrm{m} / \mathrm{sec}}=\mathrm{Nsm}^{-2}\)
JCECE-2018
Units and Measurements
139478
If the capacitance of a nanocapacitor is measured in terms of a unit ' \(u\) ' made by combining the electric charge ' \(e\) ', Bohr radius ' \(a_{0}\) ', Planck's constant ' \(h\) ' and speed of light ' \(c\) ' then
D Let ' \(u\) ' related with \(e, a_{0}, h\) and \(c\) as \([\mathrm{u}]=[\mathrm{e}]^{\mathrm{a}}\left[\mathrm{a}_{0}\right]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}}[\mathrm{c}]^{\mathrm{d}}\) Using dimension formula, \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{A}^{1} \mathrm{~T}^{1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\mathrm{c}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{d}}\) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{b}+2 \mathrm{c}+\mathrm{d}} \mathrm{T}^{\mathrm{a}-\mathrm{c}-\mathrm{d}} \mathrm{A}^{\mathrm{a}}\right]\) Comparing power \(\mathrm{a}=2, \mathrm{c}=-1\) \(\mathrm{a}-\mathrm{c}-\mathrm{d}=4\) \(d=-4+3=-1\) Putting the value of \(d\), \(b+2 c+d=-2\) \(b=1\) Hence, \(\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-1, \mathrm{~d}=-1\) Putting the value of \(a, b, c, d\) in equation (i) \(u=e^{2} a_{0}^{1} h^{-1} c^{-1}\) \(\therefore \quad \mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{a}_{0}}{\mathrm{hc}}\)
AIIMS-2016
Units and Measurements
139426
The unit of magnetic moment is
1 \(\mathrm{A}-\mathrm{m}^{2}\)
2 \(A-m\)
3 \(\mathrm{A}-\mathrm{m}^{3}\)
4 \(\mathrm{kg}-\mathrm{m}^{2}\)
Explanation:
A Magnetic moment is the product of the current flowing and area. \(\mathrm{M} =\mathrm{I} \times \mathrm{A}\) \(=\text { Ampere }-\mathrm{m}^{2}\)
139469
The velocity of a particle is given by \(v=a t^{2}+b t\) \(+c\) If \(v\) is measured in \(\mathrm{ms}^{-1}\) and \(t\) is measured in second, the unit of
1 \(a\) is \(\mathrm{ms}^{-1}\)
2 \(b\) is \(\mathrm{ms}^{-1}\)
3 \(\mathrm{c}\) is \(\mathrm{ms}^{-1}\)
4 a and b is same but that of c is different
Explanation:
C Given, Then, \(\mathrm{v}=\mathrm{at}^{2}+\mathrm{bt}+\mathrm{c}, \mathrm{v}=\mathrm{ms}^{-1} \text { and } \mathrm{t}=\text { second }\) Unit of \(\mathrm{v}=\) Unit of \(a t^{2}\) \(V=a t^{2}\) \(\mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}^{2}}=\frac{\mathrm{ms}^{-1}}{\text { Second }^{2}}=\mathrm{ms}^{-3}\) Unit of \(v=\) Unit of \(b t\) \(v=b t\) \(\mathrm{b}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{\mathrm{ms}^{-1}}{\text { Second }}=\mathrm{ms}^{-2}\) Unit of \(\mathrm{v}=\) Unit of \(\mathrm{c}\) \(\mathrm{c}=\mathrm{v}=\mathrm{ms}^{-1}\)
CG PET- 2010
Units and Measurements
139476
SI unit of coefficient of viscosity is
1 \(\mathrm{Nsm}^{-2}\)
2 \(\mathrm{Ns}^{-1} \mathrm{~m}^{2}\)
3 \(\mathrm{Nsm}^{-3}\)
4 \(\mathrm{Ns}^{-2} \mathrm{~m}\)
Explanation:
A The Co-efficient of viscosity \(\eta\) is defined as the tangential force \(F\) required to maintain a unit velocity gradient between two parallel layer of liquid of unit area \(\mathrm{A}\). Co-efficient of viscosity \((\eta)=\frac{\mathrm{Fr}}{\mathrm{AV}}\) Where, \(\mathrm{F}=\) Force \(\mathrm{A}=\) Area \(\mathrm{V}=\) Velocity \(\mathrm{r}=\) Distance between the layers \(\eta=\frac{\mathrm{N}-\mathrm{m}}{\mathrm{m}^{2} \times \mathrm{m} / \mathrm{sec}}=\mathrm{Nsm}^{-2}\)
JCECE-2018
Units and Measurements
139478
If the capacitance of a nanocapacitor is measured in terms of a unit ' \(u\) ' made by combining the electric charge ' \(e\) ', Bohr radius ' \(a_{0}\) ', Planck's constant ' \(h\) ' and speed of light ' \(c\) ' then
D Let ' \(u\) ' related with \(e, a_{0}, h\) and \(c\) as \([\mathrm{u}]=[\mathrm{e}]^{\mathrm{a}}\left[\mathrm{a}_{0}\right]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}}[\mathrm{c}]^{\mathrm{d}}\) Using dimension formula, \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{A}^{1} \mathrm{~T}^{1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\mathrm{c}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{d}}\) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{b}+2 \mathrm{c}+\mathrm{d}} \mathrm{T}^{\mathrm{a}-\mathrm{c}-\mathrm{d}} \mathrm{A}^{\mathrm{a}}\right]\) Comparing power \(\mathrm{a}=2, \mathrm{c}=-1\) \(\mathrm{a}-\mathrm{c}-\mathrm{d}=4\) \(d=-4+3=-1\) Putting the value of \(d\), \(b+2 c+d=-2\) \(b=1\) Hence, \(\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-1, \mathrm{~d}=-1\) Putting the value of \(a, b, c, d\) in equation (i) \(u=e^{2} a_{0}^{1} h^{-1} c^{-1}\) \(\therefore \quad \mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{a}_{0}}{\mathrm{hc}}\)
AIIMS-2016
Units and Measurements
139426
The unit of magnetic moment is
1 \(\mathrm{A}-\mathrm{m}^{2}\)
2 \(A-m\)
3 \(\mathrm{A}-\mathrm{m}^{3}\)
4 \(\mathrm{kg}-\mathrm{m}^{2}\)
Explanation:
A Magnetic moment is the product of the current flowing and area. \(\mathrm{M} =\mathrm{I} \times \mathrm{A}\) \(=\text { Ampere }-\mathrm{m}^{2}\)
139469
The velocity of a particle is given by \(v=a t^{2}+b t\) \(+c\) If \(v\) is measured in \(\mathrm{ms}^{-1}\) and \(t\) is measured in second, the unit of
1 \(a\) is \(\mathrm{ms}^{-1}\)
2 \(b\) is \(\mathrm{ms}^{-1}\)
3 \(\mathrm{c}\) is \(\mathrm{ms}^{-1}\)
4 a and b is same but that of c is different
Explanation:
C Given, Then, \(\mathrm{v}=\mathrm{at}^{2}+\mathrm{bt}+\mathrm{c}, \mathrm{v}=\mathrm{ms}^{-1} \text { and } \mathrm{t}=\text { second }\) Unit of \(\mathrm{v}=\) Unit of \(a t^{2}\) \(V=a t^{2}\) \(\mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}^{2}}=\frac{\mathrm{ms}^{-1}}{\text { Second }^{2}}=\mathrm{ms}^{-3}\) Unit of \(v=\) Unit of \(b t\) \(v=b t\) \(\mathrm{b}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{\mathrm{ms}^{-1}}{\text { Second }}=\mathrm{ms}^{-2}\) Unit of \(\mathrm{v}=\) Unit of \(\mathrm{c}\) \(\mathrm{c}=\mathrm{v}=\mathrm{ms}^{-1}\)
CG PET- 2010
Units and Measurements
139476
SI unit of coefficient of viscosity is
1 \(\mathrm{Nsm}^{-2}\)
2 \(\mathrm{Ns}^{-1} \mathrm{~m}^{2}\)
3 \(\mathrm{Nsm}^{-3}\)
4 \(\mathrm{Ns}^{-2} \mathrm{~m}\)
Explanation:
A The Co-efficient of viscosity \(\eta\) is defined as the tangential force \(F\) required to maintain a unit velocity gradient between two parallel layer of liquid of unit area \(\mathrm{A}\). Co-efficient of viscosity \((\eta)=\frac{\mathrm{Fr}}{\mathrm{AV}}\) Where, \(\mathrm{F}=\) Force \(\mathrm{A}=\) Area \(\mathrm{V}=\) Velocity \(\mathrm{r}=\) Distance between the layers \(\eta=\frac{\mathrm{N}-\mathrm{m}}{\mathrm{m}^{2} \times \mathrm{m} / \mathrm{sec}}=\mathrm{Nsm}^{-2}\)
JCECE-2018
Units and Measurements
139478
If the capacitance of a nanocapacitor is measured in terms of a unit ' \(u\) ' made by combining the electric charge ' \(e\) ', Bohr radius ' \(a_{0}\) ', Planck's constant ' \(h\) ' and speed of light ' \(c\) ' then
D Let ' \(u\) ' related with \(e, a_{0}, h\) and \(c\) as \([\mathrm{u}]=[\mathrm{e}]^{\mathrm{a}}\left[\mathrm{a}_{0}\right]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}}[\mathrm{c}]^{\mathrm{d}}\) Using dimension formula, \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{A}^{1} \mathrm{~T}^{1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\mathrm{c}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{d}}\) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{b}+2 \mathrm{c}+\mathrm{d}} \mathrm{T}^{\mathrm{a}-\mathrm{c}-\mathrm{d}} \mathrm{A}^{\mathrm{a}}\right]\) Comparing power \(\mathrm{a}=2, \mathrm{c}=-1\) \(\mathrm{a}-\mathrm{c}-\mathrm{d}=4\) \(d=-4+3=-1\) Putting the value of \(d\), \(b+2 c+d=-2\) \(b=1\) Hence, \(\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-1, \mathrm{~d}=-1\) Putting the value of \(a, b, c, d\) in equation (i) \(u=e^{2} a_{0}^{1} h^{-1} c^{-1}\) \(\therefore \quad \mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{a}_{0}}{\mathrm{hc}}\)
AIIMS-2016
Units and Measurements
139426
The unit of magnetic moment is
1 \(\mathrm{A}-\mathrm{m}^{2}\)
2 \(A-m\)
3 \(\mathrm{A}-\mathrm{m}^{3}\)
4 \(\mathrm{kg}-\mathrm{m}^{2}\)
Explanation:
A Magnetic moment is the product of the current flowing and area. \(\mathrm{M} =\mathrm{I} \times \mathrm{A}\) \(=\text { Ampere }-\mathrm{m}^{2}\)
139469
The velocity of a particle is given by \(v=a t^{2}+b t\) \(+c\) If \(v\) is measured in \(\mathrm{ms}^{-1}\) and \(t\) is measured in second, the unit of
1 \(a\) is \(\mathrm{ms}^{-1}\)
2 \(b\) is \(\mathrm{ms}^{-1}\)
3 \(\mathrm{c}\) is \(\mathrm{ms}^{-1}\)
4 a and b is same but that of c is different
Explanation:
C Given, Then, \(\mathrm{v}=\mathrm{at}^{2}+\mathrm{bt}+\mathrm{c}, \mathrm{v}=\mathrm{ms}^{-1} \text { and } \mathrm{t}=\text { second }\) Unit of \(\mathrm{v}=\) Unit of \(a t^{2}\) \(V=a t^{2}\) \(\mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}^{2}}=\frac{\mathrm{ms}^{-1}}{\text { Second }^{2}}=\mathrm{ms}^{-3}\) Unit of \(v=\) Unit of \(b t\) \(v=b t\) \(\mathrm{b}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{\mathrm{ms}^{-1}}{\text { Second }}=\mathrm{ms}^{-2}\) Unit of \(\mathrm{v}=\) Unit of \(\mathrm{c}\) \(\mathrm{c}=\mathrm{v}=\mathrm{ms}^{-1}\)
CG PET- 2010
Units and Measurements
139476
SI unit of coefficient of viscosity is
1 \(\mathrm{Nsm}^{-2}\)
2 \(\mathrm{Ns}^{-1} \mathrm{~m}^{2}\)
3 \(\mathrm{Nsm}^{-3}\)
4 \(\mathrm{Ns}^{-2} \mathrm{~m}\)
Explanation:
A The Co-efficient of viscosity \(\eta\) is defined as the tangential force \(F\) required to maintain a unit velocity gradient between two parallel layer of liquid of unit area \(\mathrm{A}\). Co-efficient of viscosity \((\eta)=\frac{\mathrm{Fr}}{\mathrm{AV}}\) Where, \(\mathrm{F}=\) Force \(\mathrm{A}=\) Area \(\mathrm{V}=\) Velocity \(\mathrm{r}=\) Distance between the layers \(\eta=\frac{\mathrm{N}-\mathrm{m}}{\mathrm{m}^{2} \times \mathrm{m} / \mathrm{sec}}=\mathrm{Nsm}^{-2}\)
JCECE-2018
Units and Measurements
139478
If the capacitance of a nanocapacitor is measured in terms of a unit ' \(u\) ' made by combining the electric charge ' \(e\) ', Bohr radius ' \(a_{0}\) ', Planck's constant ' \(h\) ' and speed of light ' \(c\) ' then
D Let ' \(u\) ' related with \(e, a_{0}, h\) and \(c\) as \([\mathrm{u}]=[\mathrm{e}]^{\mathrm{a}}\left[\mathrm{a}_{0}\right]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}}[\mathrm{c}]^{\mathrm{d}}\) Using dimension formula, \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{A}^{1} \mathrm{~T}^{1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\mathrm{c}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{d}}\) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{b}+2 \mathrm{c}+\mathrm{d}} \mathrm{T}^{\mathrm{a}-\mathrm{c}-\mathrm{d}} \mathrm{A}^{\mathrm{a}}\right]\) Comparing power \(\mathrm{a}=2, \mathrm{c}=-1\) \(\mathrm{a}-\mathrm{c}-\mathrm{d}=4\) \(d=-4+3=-1\) Putting the value of \(d\), \(b+2 c+d=-2\) \(b=1\) Hence, \(\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-1, \mathrm{~d}=-1\) Putting the value of \(a, b, c, d\) in equation (i) \(u=e^{2} a_{0}^{1} h^{-1} c^{-1}\) \(\therefore \quad \mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{a}_{0}}{\mathrm{hc}}\)
AIIMS-2016
Units and Measurements
139426
The unit of magnetic moment is
1 \(\mathrm{A}-\mathrm{m}^{2}\)
2 \(A-m\)
3 \(\mathrm{A}-\mathrm{m}^{3}\)
4 \(\mathrm{kg}-\mathrm{m}^{2}\)
Explanation:
A Magnetic moment is the product of the current flowing and area. \(\mathrm{M} =\mathrm{I} \times \mathrm{A}\) \(=\text { Ampere }-\mathrm{m}^{2}\)
