367323
Statement A : Pressure cannot be subtracted from pressure gradient. Statement B : Pressure and pressure gradient have different dimensions.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Pressure and pressure graient have different dimensions. Hence pressure can not be subtracted from pressure gradient.So option (3) is correct.
PHXI02:UNITS AND MEASUREMENTS
367324
In the relation \(x=\cos (\omega t+k x)\), the dimensions of \(\omega\) are
1 \(\left[M^{0} L T^{-1}\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L^{0} T^{-1}\right]\)
4 \(\left[M^{0} L T\right] \mathrm{s}\)
Explanation:
\(\because x=\cos (\omega t+k x)\) \(\therefore(\omega t+k x)\) is an angle whose dimensions will be \(\left[M^{0} L^{0} T^{0}\right]\). \(\therefore\) Dimensions of \(\omega t=\left[M^{0} L^{0} T^{0}\right]\) \(\therefore\) Dimensions of \(\omega=\left[M^{0} L^{0} T^{-1}\right]\).
PHXI02:UNITS AND MEASUREMENTS
367325
\(A,\) \(B,\) \(C\) and \(D\) are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation \(AD = C\,\,\ln \,\,(BD)\) holds true. Then which of the combination is not a meaningful quantity?
1 \(\dfrac{(A-C)}{D}\)
2 \(A^{2}-B^{2} C^{2}\)
3 \(\dfrac{A}{B}-C\)
4 \(\dfrac{C}{B D}-\dfrac{A D^{2}}{C}\)
Explanation:
Dimension of \(A \ne \) dimension of \((C)\) Hence \(A - C\) is not possible.
JEE - 2016
PHXI02:UNITS AND MEASUREMENTS
367326
If \(F\) denotes force and \(t\) time, then in the equation \(F=a t^{-1}+b t^{2}\), dimensions of \(a\) and \(b\) respectively are
1 \(\left[L T^{-4}\right]\) and \(\left[L T^{-1}\right]\)
2 \(\left[L T^{-1}\right]\) and \(\left[L T^{-4}\right]\)
3 \(\left[M L T^{-4}\right]\) and \(\left[M L T^{-1}\right]\)
4 \(\left[M L T^{-1}\right]\) and \(\left[M L T^{-4}\right]\)
Explanation:
\(F=a t^{-1}+b t^{2}\) \(\begin{gathered}\therefore[a]=[F][t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\{[b]=\dfrac{[F]}{\left[t^{2}\right]}=\dfrac{\left[M L T^{-2}\right]}{[T]}=\left[M L T^{-4}\right] .}\end{gathered}\)
PHXI02:UNITS AND MEASUREMENTS
367327
If \(x=a t+b t^{2}\), where \(x\) is the distance travelled by the body in kilometre, while \(t\) is the time in seconds, then the units of \(b\) are
1 \(km{\rm{/}}s\)
2 \(km - s\)
3 \(km{\rm{/}}{s^2}\)
4 \(km - {s^2}\)
Explanation:
From the principle of dimensional homogeneity \([x]=\left[b t^{2}\right] \Rightarrow[b]=\left[\dfrac{x}{t^{2}}\right]\) \(\therefore\) Unit of \(b = km{\rm{/}}{s^2}\).
367323
Statement A : Pressure cannot be subtracted from pressure gradient. Statement B : Pressure and pressure gradient have different dimensions.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Pressure and pressure graient have different dimensions. Hence pressure can not be subtracted from pressure gradient.So option (3) is correct.
PHXI02:UNITS AND MEASUREMENTS
367324
In the relation \(x=\cos (\omega t+k x)\), the dimensions of \(\omega\) are
1 \(\left[M^{0} L T^{-1}\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L^{0} T^{-1}\right]\)
4 \(\left[M^{0} L T\right] \mathrm{s}\)
Explanation:
\(\because x=\cos (\omega t+k x)\) \(\therefore(\omega t+k x)\) is an angle whose dimensions will be \(\left[M^{0} L^{0} T^{0}\right]\). \(\therefore\) Dimensions of \(\omega t=\left[M^{0} L^{0} T^{0}\right]\) \(\therefore\) Dimensions of \(\omega=\left[M^{0} L^{0} T^{-1}\right]\).
PHXI02:UNITS AND MEASUREMENTS
367325
\(A,\) \(B,\) \(C\) and \(D\) are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation \(AD = C\,\,\ln \,\,(BD)\) holds true. Then which of the combination is not a meaningful quantity?
1 \(\dfrac{(A-C)}{D}\)
2 \(A^{2}-B^{2} C^{2}\)
3 \(\dfrac{A}{B}-C\)
4 \(\dfrac{C}{B D}-\dfrac{A D^{2}}{C}\)
Explanation:
Dimension of \(A \ne \) dimension of \((C)\) Hence \(A - C\) is not possible.
JEE - 2016
PHXI02:UNITS AND MEASUREMENTS
367326
If \(F\) denotes force and \(t\) time, then in the equation \(F=a t^{-1}+b t^{2}\), dimensions of \(a\) and \(b\) respectively are
1 \(\left[L T^{-4}\right]\) and \(\left[L T^{-1}\right]\)
2 \(\left[L T^{-1}\right]\) and \(\left[L T^{-4}\right]\)
3 \(\left[M L T^{-4}\right]\) and \(\left[M L T^{-1}\right]\)
4 \(\left[M L T^{-1}\right]\) and \(\left[M L T^{-4}\right]\)
Explanation:
\(F=a t^{-1}+b t^{2}\) \(\begin{gathered}\therefore[a]=[F][t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\{[b]=\dfrac{[F]}{\left[t^{2}\right]}=\dfrac{\left[M L T^{-2}\right]}{[T]}=\left[M L T^{-4}\right] .}\end{gathered}\)
PHXI02:UNITS AND MEASUREMENTS
367327
If \(x=a t+b t^{2}\), where \(x\) is the distance travelled by the body in kilometre, while \(t\) is the time in seconds, then the units of \(b\) are
1 \(km{\rm{/}}s\)
2 \(km - s\)
3 \(km{\rm{/}}{s^2}\)
4 \(km - {s^2}\)
Explanation:
From the principle of dimensional homogeneity \([x]=\left[b t^{2}\right] \Rightarrow[b]=\left[\dfrac{x}{t^{2}}\right]\) \(\therefore\) Unit of \(b = km{\rm{/}}{s^2}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI02:UNITS AND MEASUREMENTS
367323
Statement A : Pressure cannot be subtracted from pressure gradient. Statement B : Pressure and pressure gradient have different dimensions.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Pressure and pressure graient have different dimensions. Hence pressure can not be subtracted from pressure gradient.So option (3) is correct.
PHXI02:UNITS AND MEASUREMENTS
367324
In the relation \(x=\cos (\omega t+k x)\), the dimensions of \(\omega\) are
1 \(\left[M^{0} L T^{-1}\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L^{0} T^{-1}\right]\)
4 \(\left[M^{0} L T\right] \mathrm{s}\)
Explanation:
\(\because x=\cos (\omega t+k x)\) \(\therefore(\omega t+k x)\) is an angle whose dimensions will be \(\left[M^{0} L^{0} T^{0}\right]\). \(\therefore\) Dimensions of \(\omega t=\left[M^{0} L^{0} T^{0}\right]\) \(\therefore\) Dimensions of \(\omega=\left[M^{0} L^{0} T^{-1}\right]\).
PHXI02:UNITS AND MEASUREMENTS
367325
\(A,\) \(B,\) \(C\) and \(D\) are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation \(AD = C\,\,\ln \,\,(BD)\) holds true. Then which of the combination is not a meaningful quantity?
1 \(\dfrac{(A-C)}{D}\)
2 \(A^{2}-B^{2} C^{2}\)
3 \(\dfrac{A}{B}-C\)
4 \(\dfrac{C}{B D}-\dfrac{A D^{2}}{C}\)
Explanation:
Dimension of \(A \ne \) dimension of \((C)\) Hence \(A - C\) is not possible.
JEE - 2016
PHXI02:UNITS AND MEASUREMENTS
367326
If \(F\) denotes force and \(t\) time, then in the equation \(F=a t^{-1}+b t^{2}\), dimensions of \(a\) and \(b\) respectively are
1 \(\left[L T^{-4}\right]\) and \(\left[L T^{-1}\right]\)
2 \(\left[L T^{-1}\right]\) and \(\left[L T^{-4}\right]\)
3 \(\left[M L T^{-4}\right]\) and \(\left[M L T^{-1}\right]\)
4 \(\left[M L T^{-1}\right]\) and \(\left[M L T^{-4}\right]\)
Explanation:
\(F=a t^{-1}+b t^{2}\) \(\begin{gathered}\therefore[a]=[F][t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\{[b]=\dfrac{[F]}{\left[t^{2}\right]}=\dfrac{\left[M L T^{-2}\right]}{[T]}=\left[M L T^{-4}\right] .}\end{gathered}\)
PHXI02:UNITS AND MEASUREMENTS
367327
If \(x=a t+b t^{2}\), where \(x\) is the distance travelled by the body in kilometre, while \(t\) is the time in seconds, then the units of \(b\) are
1 \(km{\rm{/}}s\)
2 \(km - s\)
3 \(km{\rm{/}}{s^2}\)
4 \(km - {s^2}\)
Explanation:
From the principle of dimensional homogeneity \([x]=\left[b t^{2}\right] \Rightarrow[b]=\left[\dfrac{x}{t^{2}}\right]\) \(\therefore\) Unit of \(b = km{\rm{/}}{s^2}\).
367323
Statement A : Pressure cannot be subtracted from pressure gradient. Statement B : Pressure and pressure gradient have different dimensions.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Pressure and pressure graient have different dimensions. Hence pressure can not be subtracted from pressure gradient.So option (3) is correct.
PHXI02:UNITS AND MEASUREMENTS
367324
In the relation \(x=\cos (\omega t+k x)\), the dimensions of \(\omega\) are
1 \(\left[M^{0} L T^{-1}\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L^{0} T^{-1}\right]\)
4 \(\left[M^{0} L T\right] \mathrm{s}\)
Explanation:
\(\because x=\cos (\omega t+k x)\) \(\therefore(\omega t+k x)\) is an angle whose dimensions will be \(\left[M^{0} L^{0} T^{0}\right]\). \(\therefore\) Dimensions of \(\omega t=\left[M^{0} L^{0} T^{0}\right]\) \(\therefore\) Dimensions of \(\omega=\left[M^{0} L^{0} T^{-1}\right]\).
PHXI02:UNITS AND MEASUREMENTS
367325
\(A,\) \(B,\) \(C\) and \(D\) are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation \(AD = C\,\,\ln \,\,(BD)\) holds true. Then which of the combination is not a meaningful quantity?
1 \(\dfrac{(A-C)}{D}\)
2 \(A^{2}-B^{2} C^{2}\)
3 \(\dfrac{A}{B}-C\)
4 \(\dfrac{C}{B D}-\dfrac{A D^{2}}{C}\)
Explanation:
Dimension of \(A \ne \) dimension of \((C)\) Hence \(A - C\) is not possible.
JEE - 2016
PHXI02:UNITS AND MEASUREMENTS
367326
If \(F\) denotes force and \(t\) time, then in the equation \(F=a t^{-1}+b t^{2}\), dimensions of \(a\) and \(b\) respectively are
1 \(\left[L T^{-4}\right]\) and \(\left[L T^{-1}\right]\)
2 \(\left[L T^{-1}\right]\) and \(\left[L T^{-4}\right]\)
3 \(\left[M L T^{-4}\right]\) and \(\left[M L T^{-1}\right]\)
4 \(\left[M L T^{-1}\right]\) and \(\left[M L T^{-4}\right]\)
Explanation:
\(F=a t^{-1}+b t^{2}\) \(\begin{gathered}\therefore[a]=[F][t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\{[b]=\dfrac{[F]}{\left[t^{2}\right]}=\dfrac{\left[M L T^{-2}\right]}{[T]}=\left[M L T^{-4}\right] .}\end{gathered}\)
PHXI02:UNITS AND MEASUREMENTS
367327
If \(x=a t+b t^{2}\), where \(x\) is the distance travelled by the body in kilometre, while \(t\) is the time in seconds, then the units of \(b\) are
1 \(km{\rm{/}}s\)
2 \(km - s\)
3 \(km{\rm{/}}{s^2}\)
4 \(km - {s^2}\)
Explanation:
From the principle of dimensional homogeneity \([x]=\left[b t^{2}\right] \Rightarrow[b]=\left[\dfrac{x}{t^{2}}\right]\) \(\therefore\) Unit of \(b = km{\rm{/}}{s^2}\).
367323
Statement A : Pressure cannot be subtracted from pressure gradient. Statement B : Pressure and pressure gradient have different dimensions.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Pressure and pressure graient have different dimensions. Hence pressure can not be subtracted from pressure gradient.So option (3) is correct.
PHXI02:UNITS AND MEASUREMENTS
367324
In the relation \(x=\cos (\omega t+k x)\), the dimensions of \(\omega\) are
1 \(\left[M^{0} L T^{-1}\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L^{0} T^{-1}\right]\)
4 \(\left[M^{0} L T\right] \mathrm{s}\)
Explanation:
\(\because x=\cos (\omega t+k x)\) \(\therefore(\omega t+k x)\) is an angle whose dimensions will be \(\left[M^{0} L^{0} T^{0}\right]\). \(\therefore\) Dimensions of \(\omega t=\left[M^{0} L^{0} T^{0}\right]\) \(\therefore\) Dimensions of \(\omega=\left[M^{0} L^{0} T^{-1}\right]\).
PHXI02:UNITS AND MEASUREMENTS
367325
\(A,\) \(B,\) \(C\) and \(D\) are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation \(AD = C\,\,\ln \,\,(BD)\) holds true. Then which of the combination is not a meaningful quantity?
1 \(\dfrac{(A-C)}{D}\)
2 \(A^{2}-B^{2} C^{2}\)
3 \(\dfrac{A}{B}-C\)
4 \(\dfrac{C}{B D}-\dfrac{A D^{2}}{C}\)
Explanation:
Dimension of \(A \ne \) dimension of \((C)\) Hence \(A - C\) is not possible.
JEE - 2016
PHXI02:UNITS AND MEASUREMENTS
367326
If \(F\) denotes force and \(t\) time, then in the equation \(F=a t^{-1}+b t^{2}\), dimensions of \(a\) and \(b\) respectively are
1 \(\left[L T^{-4}\right]\) and \(\left[L T^{-1}\right]\)
2 \(\left[L T^{-1}\right]\) and \(\left[L T^{-4}\right]\)
3 \(\left[M L T^{-4}\right]\) and \(\left[M L T^{-1}\right]\)
4 \(\left[M L T^{-1}\right]\) and \(\left[M L T^{-4}\right]\)
Explanation:
\(F=a t^{-1}+b t^{2}\) \(\begin{gathered}\therefore[a]=[F][t]=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\{[b]=\dfrac{[F]}{\left[t^{2}\right]}=\dfrac{\left[M L T^{-2}\right]}{[T]}=\left[M L T^{-4}\right] .}\end{gathered}\)
PHXI02:UNITS AND MEASUREMENTS
367327
If \(x=a t+b t^{2}\), where \(x\) is the distance travelled by the body in kilometre, while \(t\) is the time in seconds, then the units of \(b\) are
1 \(km{\rm{/}}s\)
2 \(km - s\)
3 \(km{\rm{/}}{s^2}\)
4 \(km - {s^2}\)
Explanation:
From the principle of dimensional homogeneity \([x]=\left[b t^{2}\right] \Rightarrow[b]=\left[\dfrac{x}{t^{2}}\right]\) \(\therefore\) Unit of \(b = km{\rm{/}}{s^2}\).