139509
The equation of state of some gases can be expressed as \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) where, \(p\) is the pressure, \(V\) the volume, \(T\) the absolute temperature and \(a\) and \(b\) are constants. The dimensional formula of \(a\) is
A According to the principle of dimensional homogeneity the dimensions of each the terms of a dimensional equation on both sides are the same. So, dimension of \(\mathrm{p}\) and \(\frac{\mathrm{a}}{\mathrm{V}^{2}}\) will be same \(\mathrm{p}=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right]\) \(\mathrm{a}=[\mathrm{p}]\left[\mathrm{V}^{2}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]\)
UPSEE-2002
Units and Measurements
139510
The dimensions of universal gravitational constant are-
C Gravitational force acting between two body is- \(\mathrm{F} =\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) \(\therefore \quad \mathrm{G} =\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{~m}_{1} \cdot \mathrm{m}_{2}}\) Where, \(G=\) Universal gravitational constant \(\mathrm{F}=\) Gravitational force \(\mathrm{m}_{1}=\) mass of first body \(\mathrm{m}_{2}=\) mass of second body \(\mathrm{r}=\) Distance between two body. \(\therefore\) Unit of Universal gravitational constant- \(\Rightarrow\) Unit of \(\mathrm{G}=\frac{\text { Newton. } \mathrm{m}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}=\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\) Dimension of \(\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Kerala PMT-2012
Units and Measurements
139513
If the force is given by \(F=a t+b t^{2}\) with \(t\) as time. The dimensions of \(a\) and \(b\) are
1 \(\left[\mathrm{MLT}^{-4}\right]\) and \(\left[\mathrm{MLT}^{-2}\right]\)
2 \(\left[\mathrm{MLT}^{-3}\right]\) and \(\left[\mathrm{MLT}^{-4}\right]\)
3 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
4 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]\)
Explanation:
B Given, \(\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}\) Dimension of \(\mathrm{F}=\) Dimension of at \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{a}[\mathrm{T}]}\) \(\mathrm{a}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}\) \(\mathrm{a}=\left[\mathrm{MLT}^{-3}\right]\) Dimension of \(\mathrm{F}=\) Dimension of \(\mathrm{bt}^{2}\) \(\mathrm{F}=\mathrm{bt}^{2}\) \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{b}[\mathrm{T}]^{2}}\) \(\mathrm{~b}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^{2}\right]}\) \(\mathrm{b}=\left[\mathrm{MLT}^{-4}\right]\)
C We know Angular momentum \((\mathrm{L})=\mathrm{mvr}\). \(\text { where }= \mathrm{m}=\operatorname{mass}(\mathrm{kg})\) \(\mathrm{V}=\operatorname{velocity}(\mathrm{m} / \mathrm{s})\) \(\mathrm{r}=\operatorname{radius}(\mathrm{m})\) \(\mathrm{L}=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(\mathrm{L}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
IIT-1983
Units and Measurements
139516
The speed of light (c), gravitation constant (G), and Plank's constant \((h)\) are taken as the fundamental units in a system. The dimension of time in this new system should be:
1 \(\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}\right]\)
2 \(\left[G^{-1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
3 \(\left[G^{1 / 2} h^{1 / 2} c^{-3 / 2}\right]\)
4 \(\left[G^{1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
Explanation:
A dimension \({[\mathrm{C}]=\mathrm{LT}^{-1}}\) \({[\mathrm{G}]=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}}\) \({[\mathrm{~h}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\) Let \(\quad t \propto c^{x} G^{y} h^{z}\) \(\mathrm{T} =\left(\mathrm{LT}^{-1}\right)^{\mathrm{x}}\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)^{\mathrm{y}}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\mathrm{z}}\) \(=[\mathrm{M}]^{-\mathrm{y}+\mathrm{z}}[\mathrm{L}]^{\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}}[\mathrm{T}]^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}\) By equating \(-y+z=0\) \(x+3 y+2 z=0\) \(-x-2 y-z=1\) By Solving eq \({ }^{\mathrm{n}}\) (i), (ii) \& (iii) We get, \(\mathrm{x}=\frac{-5}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}\) \([t]=\left[C^{-5 / 2} G^{1 / 2} h^{1 / 2}\right]\)
139509
The equation of state of some gases can be expressed as \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) where, \(p\) is the pressure, \(V\) the volume, \(T\) the absolute temperature and \(a\) and \(b\) are constants. The dimensional formula of \(a\) is
A According to the principle of dimensional homogeneity the dimensions of each the terms of a dimensional equation on both sides are the same. So, dimension of \(\mathrm{p}\) and \(\frac{\mathrm{a}}{\mathrm{V}^{2}}\) will be same \(\mathrm{p}=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right]\) \(\mathrm{a}=[\mathrm{p}]\left[\mathrm{V}^{2}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]\)
UPSEE-2002
Units and Measurements
139510
The dimensions of universal gravitational constant are-
C Gravitational force acting between two body is- \(\mathrm{F} =\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) \(\therefore \quad \mathrm{G} =\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{~m}_{1} \cdot \mathrm{m}_{2}}\) Where, \(G=\) Universal gravitational constant \(\mathrm{F}=\) Gravitational force \(\mathrm{m}_{1}=\) mass of first body \(\mathrm{m}_{2}=\) mass of second body \(\mathrm{r}=\) Distance between two body. \(\therefore\) Unit of Universal gravitational constant- \(\Rightarrow\) Unit of \(\mathrm{G}=\frac{\text { Newton. } \mathrm{m}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}=\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\) Dimension of \(\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Kerala PMT-2012
Units and Measurements
139513
If the force is given by \(F=a t+b t^{2}\) with \(t\) as time. The dimensions of \(a\) and \(b\) are
1 \(\left[\mathrm{MLT}^{-4}\right]\) and \(\left[\mathrm{MLT}^{-2}\right]\)
2 \(\left[\mathrm{MLT}^{-3}\right]\) and \(\left[\mathrm{MLT}^{-4}\right]\)
3 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
4 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]\)
Explanation:
B Given, \(\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}\) Dimension of \(\mathrm{F}=\) Dimension of at \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{a}[\mathrm{T}]}\) \(\mathrm{a}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}\) \(\mathrm{a}=\left[\mathrm{MLT}^{-3}\right]\) Dimension of \(\mathrm{F}=\) Dimension of \(\mathrm{bt}^{2}\) \(\mathrm{F}=\mathrm{bt}^{2}\) \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{b}[\mathrm{T}]^{2}}\) \(\mathrm{~b}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^{2}\right]}\) \(\mathrm{b}=\left[\mathrm{MLT}^{-4}\right]\)
C We know Angular momentum \((\mathrm{L})=\mathrm{mvr}\). \(\text { where }= \mathrm{m}=\operatorname{mass}(\mathrm{kg})\) \(\mathrm{V}=\operatorname{velocity}(\mathrm{m} / \mathrm{s})\) \(\mathrm{r}=\operatorname{radius}(\mathrm{m})\) \(\mathrm{L}=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(\mathrm{L}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
IIT-1983
Units and Measurements
139516
The speed of light (c), gravitation constant (G), and Plank's constant \((h)\) are taken as the fundamental units in a system. The dimension of time in this new system should be:
1 \(\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}\right]\)
2 \(\left[G^{-1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
3 \(\left[G^{1 / 2} h^{1 / 2} c^{-3 / 2}\right]\)
4 \(\left[G^{1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
Explanation:
A dimension \({[\mathrm{C}]=\mathrm{LT}^{-1}}\) \({[\mathrm{G}]=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}}\) \({[\mathrm{~h}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\) Let \(\quad t \propto c^{x} G^{y} h^{z}\) \(\mathrm{T} =\left(\mathrm{LT}^{-1}\right)^{\mathrm{x}}\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)^{\mathrm{y}}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\mathrm{z}}\) \(=[\mathrm{M}]^{-\mathrm{y}+\mathrm{z}}[\mathrm{L}]^{\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}}[\mathrm{T}]^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}\) By equating \(-y+z=0\) \(x+3 y+2 z=0\) \(-x-2 y-z=1\) By Solving eq \({ }^{\mathrm{n}}\) (i), (ii) \& (iii) We get, \(\mathrm{x}=\frac{-5}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}\) \([t]=\left[C^{-5 / 2} G^{1 / 2} h^{1 / 2}\right]\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Units and Measurements
139509
The equation of state of some gases can be expressed as \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) where, \(p\) is the pressure, \(V\) the volume, \(T\) the absolute temperature and \(a\) and \(b\) are constants. The dimensional formula of \(a\) is
A According to the principle of dimensional homogeneity the dimensions of each the terms of a dimensional equation on both sides are the same. So, dimension of \(\mathrm{p}\) and \(\frac{\mathrm{a}}{\mathrm{V}^{2}}\) will be same \(\mathrm{p}=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right]\) \(\mathrm{a}=[\mathrm{p}]\left[\mathrm{V}^{2}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]\)
UPSEE-2002
Units and Measurements
139510
The dimensions of universal gravitational constant are-
C Gravitational force acting between two body is- \(\mathrm{F} =\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) \(\therefore \quad \mathrm{G} =\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{~m}_{1} \cdot \mathrm{m}_{2}}\) Where, \(G=\) Universal gravitational constant \(\mathrm{F}=\) Gravitational force \(\mathrm{m}_{1}=\) mass of first body \(\mathrm{m}_{2}=\) mass of second body \(\mathrm{r}=\) Distance between two body. \(\therefore\) Unit of Universal gravitational constant- \(\Rightarrow\) Unit of \(\mathrm{G}=\frac{\text { Newton. } \mathrm{m}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}=\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\) Dimension of \(\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Kerala PMT-2012
Units and Measurements
139513
If the force is given by \(F=a t+b t^{2}\) with \(t\) as time. The dimensions of \(a\) and \(b\) are
1 \(\left[\mathrm{MLT}^{-4}\right]\) and \(\left[\mathrm{MLT}^{-2}\right]\)
2 \(\left[\mathrm{MLT}^{-3}\right]\) and \(\left[\mathrm{MLT}^{-4}\right]\)
3 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
4 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]\)
Explanation:
B Given, \(\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}\) Dimension of \(\mathrm{F}=\) Dimension of at \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{a}[\mathrm{T}]}\) \(\mathrm{a}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}\) \(\mathrm{a}=\left[\mathrm{MLT}^{-3}\right]\) Dimension of \(\mathrm{F}=\) Dimension of \(\mathrm{bt}^{2}\) \(\mathrm{F}=\mathrm{bt}^{2}\) \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{b}[\mathrm{T}]^{2}}\) \(\mathrm{~b}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^{2}\right]}\) \(\mathrm{b}=\left[\mathrm{MLT}^{-4}\right]\)
C We know Angular momentum \((\mathrm{L})=\mathrm{mvr}\). \(\text { where }= \mathrm{m}=\operatorname{mass}(\mathrm{kg})\) \(\mathrm{V}=\operatorname{velocity}(\mathrm{m} / \mathrm{s})\) \(\mathrm{r}=\operatorname{radius}(\mathrm{m})\) \(\mathrm{L}=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(\mathrm{L}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
IIT-1983
Units and Measurements
139516
The speed of light (c), gravitation constant (G), and Plank's constant \((h)\) are taken as the fundamental units in a system. The dimension of time in this new system should be:
1 \(\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}\right]\)
2 \(\left[G^{-1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
3 \(\left[G^{1 / 2} h^{1 / 2} c^{-3 / 2}\right]\)
4 \(\left[G^{1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
Explanation:
A dimension \({[\mathrm{C}]=\mathrm{LT}^{-1}}\) \({[\mathrm{G}]=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}}\) \({[\mathrm{~h}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\) Let \(\quad t \propto c^{x} G^{y} h^{z}\) \(\mathrm{T} =\left(\mathrm{LT}^{-1}\right)^{\mathrm{x}}\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)^{\mathrm{y}}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\mathrm{z}}\) \(=[\mathrm{M}]^{-\mathrm{y}+\mathrm{z}}[\mathrm{L}]^{\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}}[\mathrm{T}]^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}\) By equating \(-y+z=0\) \(x+3 y+2 z=0\) \(-x-2 y-z=1\) By Solving eq \({ }^{\mathrm{n}}\) (i), (ii) \& (iii) We get, \(\mathrm{x}=\frac{-5}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}\) \([t]=\left[C^{-5 / 2} G^{1 / 2} h^{1 / 2}\right]\)
139509
The equation of state of some gases can be expressed as \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) where, \(p\) is the pressure, \(V\) the volume, \(T\) the absolute temperature and \(a\) and \(b\) are constants. The dimensional formula of \(a\) is
A According to the principle of dimensional homogeneity the dimensions of each the terms of a dimensional equation on both sides are the same. So, dimension of \(\mathrm{p}\) and \(\frac{\mathrm{a}}{\mathrm{V}^{2}}\) will be same \(\mathrm{p}=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right]\) \(\mathrm{a}=[\mathrm{p}]\left[\mathrm{V}^{2}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]\)
UPSEE-2002
Units and Measurements
139510
The dimensions of universal gravitational constant are-
C Gravitational force acting between two body is- \(\mathrm{F} =\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) \(\therefore \quad \mathrm{G} =\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{~m}_{1} \cdot \mathrm{m}_{2}}\) Where, \(G=\) Universal gravitational constant \(\mathrm{F}=\) Gravitational force \(\mathrm{m}_{1}=\) mass of first body \(\mathrm{m}_{2}=\) mass of second body \(\mathrm{r}=\) Distance between two body. \(\therefore\) Unit of Universal gravitational constant- \(\Rightarrow\) Unit of \(\mathrm{G}=\frac{\text { Newton. } \mathrm{m}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}=\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\) Dimension of \(\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Kerala PMT-2012
Units and Measurements
139513
If the force is given by \(F=a t+b t^{2}\) with \(t\) as time. The dimensions of \(a\) and \(b\) are
1 \(\left[\mathrm{MLT}^{-4}\right]\) and \(\left[\mathrm{MLT}^{-2}\right]\)
2 \(\left[\mathrm{MLT}^{-3}\right]\) and \(\left[\mathrm{MLT}^{-4}\right]\)
3 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
4 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]\)
Explanation:
B Given, \(\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}\) Dimension of \(\mathrm{F}=\) Dimension of at \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{a}[\mathrm{T}]}\) \(\mathrm{a}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}\) \(\mathrm{a}=\left[\mathrm{MLT}^{-3}\right]\) Dimension of \(\mathrm{F}=\) Dimension of \(\mathrm{bt}^{2}\) \(\mathrm{F}=\mathrm{bt}^{2}\) \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{b}[\mathrm{T}]^{2}}\) \(\mathrm{~b}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^{2}\right]}\) \(\mathrm{b}=\left[\mathrm{MLT}^{-4}\right]\)
C We know Angular momentum \((\mathrm{L})=\mathrm{mvr}\). \(\text { where }= \mathrm{m}=\operatorname{mass}(\mathrm{kg})\) \(\mathrm{V}=\operatorname{velocity}(\mathrm{m} / \mathrm{s})\) \(\mathrm{r}=\operatorname{radius}(\mathrm{m})\) \(\mathrm{L}=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(\mathrm{L}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
IIT-1983
Units and Measurements
139516
The speed of light (c), gravitation constant (G), and Plank's constant \((h)\) are taken as the fundamental units in a system. The dimension of time in this new system should be:
1 \(\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}\right]\)
2 \(\left[G^{-1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
3 \(\left[G^{1 / 2} h^{1 / 2} c^{-3 / 2}\right]\)
4 \(\left[G^{1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
Explanation:
A dimension \({[\mathrm{C}]=\mathrm{LT}^{-1}}\) \({[\mathrm{G}]=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}}\) \({[\mathrm{~h}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\) Let \(\quad t \propto c^{x} G^{y} h^{z}\) \(\mathrm{T} =\left(\mathrm{LT}^{-1}\right)^{\mathrm{x}}\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)^{\mathrm{y}}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\mathrm{z}}\) \(=[\mathrm{M}]^{-\mathrm{y}+\mathrm{z}}[\mathrm{L}]^{\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}}[\mathrm{T}]^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}\) By equating \(-y+z=0\) \(x+3 y+2 z=0\) \(-x-2 y-z=1\) By Solving eq \({ }^{\mathrm{n}}\) (i), (ii) \& (iii) We get, \(\mathrm{x}=\frac{-5}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}\) \([t]=\left[C^{-5 / 2} G^{1 / 2} h^{1 / 2}\right]\)
139509
The equation of state of some gases can be expressed as \(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\) where, \(p\) is the pressure, \(V\) the volume, \(T\) the absolute temperature and \(a\) and \(b\) are constants. The dimensional formula of \(a\) is
A According to the principle of dimensional homogeneity the dimensions of each the terms of a dimensional equation on both sides are the same. So, dimension of \(\mathrm{p}\) and \(\frac{\mathrm{a}}{\mathrm{V}^{2}}\) will be same \(\mathrm{p}=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right]\) \(\mathrm{a}=[\mathrm{p}]\left[\mathrm{V}^{2}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]\) \(\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]\)
UPSEE-2002
Units and Measurements
139510
The dimensions of universal gravitational constant are-
C Gravitational force acting between two body is- \(\mathrm{F} =\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) \(\therefore \quad \mathrm{G} =\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{~m}_{1} \cdot \mathrm{m}_{2}}\) Where, \(G=\) Universal gravitational constant \(\mathrm{F}=\) Gravitational force \(\mathrm{m}_{1}=\) mass of first body \(\mathrm{m}_{2}=\) mass of second body \(\mathrm{r}=\) Distance between two body. \(\therefore\) Unit of Universal gravitational constant- \(\Rightarrow\) Unit of \(\mathrm{G}=\frac{\text { Newton. } \mathrm{m}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}=\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\) Dimension of \(\frac{\mathrm{N} \cdot \mathrm{m}^{2}}{(\mathrm{Kg})^{2}}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}\) \(\Rightarrow\) Dimension of \(\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Kerala PMT-2012
Units and Measurements
139513
If the force is given by \(F=a t+b t^{2}\) with \(t\) as time. The dimensions of \(a\) and \(b\) are
1 \(\left[\mathrm{MLT}^{-4}\right]\) and \(\left[\mathrm{MLT}^{-2}\right]\)
2 \(\left[\mathrm{MLT}^{-3}\right]\) and \(\left[\mathrm{MLT}^{-4}\right]\)
3 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
4 \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\) and \(\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]\)
Explanation:
B Given, \(\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}\) Dimension of \(\mathrm{F}=\) Dimension of at \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{a}[\mathrm{T}]}\) \(\mathrm{a}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}\) \(\mathrm{a}=\left[\mathrm{MLT}^{-3}\right]\) Dimension of \(\mathrm{F}=\) Dimension of \(\mathrm{bt}^{2}\) \(\mathrm{F}=\mathrm{bt}^{2}\) \({\left[\mathrm{MLT}^{-2}\right]=\mathrm{b}[\mathrm{T}]^{2}}\) \(\mathrm{~b}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^{2}\right]}\) \(\mathrm{b}=\left[\mathrm{MLT}^{-4}\right]\)
C We know Angular momentum \((\mathrm{L})=\mathrm{mvr}\). \(\text { where }= \mathrm{m}=\operatorname{mass}(\mathrm{kg})\) \(\mathrm{V}=\operatorname{velocity}(\mathrm{m} / \mathrm{s})\) \(\mathrm{r}=\operatorname{radius}(\mathrm{m})\) \(\mathrm{L}=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]\) \(\mathrm{L}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
IIT-1983
Units and Measurements
139516
The speed of light (c), gravitation constant (G), and Plank's constant \((h)\) are taken as the fundamental units in a system. The dimension of time in this new system should be:
1 \(\left[G^{1 / 2} h^{1 / 2} c^{-5 / 2}\right]\)
2 \(\left[G^{-1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
3 \(\left[G^{1 / 2} h^{1 / 2} c^{-3 / 2}\right]\)
4 \(\left[G^{1 / 2} h^{1 / 2} c^{1 / 2}\right]\)
Explanation:
A dimension \({[\mathrm{C}]=\mathrm{LT}^{-1}}\) \({[\mathrm{G}]=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}}\) \({[\mathrm{~h}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\) Let \(\quad t \propto c^{x} G^{y} h^{z}\) \(\mathrm{T} =\left(\mathrm{LT}^{-1}\right)^{\mathrm{x}}\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)^{\mathrm{y}}\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\mathrm{z}}\) \(=[\mathrm{M}]^{-\mathrm{y}+\mathrm{z}}[\mathrm{L}]^{\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}}[\mathrm{T}]^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}\) By equating \(-y+z=0\) \(x+3 y+2 z=0\) \(-x-2 y-z=1\) By Solving eq \({ }^{\mathrm{n}}\) (i), (ii) \& (iii) We get, \(\mathrm{x}=\frac{-5}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}\) \([t]=\left[C^{-5 / 2} G^{1 / 2} h^{1 / 2}\right]\)