NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI02:UNITS AND MEASUREMENTS
367289
In relation \(P = \frac{a}{b}{e^{ - \left( {aE/t} \right)}}\) \(P\) represents number, \(E\) is energy and \(t\) is time. The dimension of \(a\) and \(b\) is
As \(P\) is a number then \(a\) and \(b\) have same units \(\frac{{aE}}{t} = {M^o}{L^o}{T^o}\) \(\frac{{a\left( {M{L^2}{T^{ - 2}}} \right)}}{T} = {M^o}{L^o}{T^o}\) \(a = ({M^{ - 1}}{L^{ - 2}}{T^3}) = b\)
PHXI02:UNITS AND MEASUREMENTS
367290
Find the dimensional formula of surface tension \(S\),equation involving the surface tension \(S = \frac{{\rho grh}}{2}\)
367291
In the relation, \({P=\dfrac{\alpha}{\beta} e^{-\dfrac{\alpha}{k \theta}} P}\) is pressure, \({Z}\) is the distance, \({k}\) is Boltzmann constant, and \({\theta}\) is the temperature. The dimensional formula of \({\beta}\) will be
1 \({\left[M^{0} L^{2} T^{0}\right]}\)
2 \({\left[M^{1} L^{2} T^{1}\right]}\)
3 \({\left[M^{1} L^{0} T^{-1}\right]}\)
4 \({\left[M^{1} L^{2} T^{-1}\right]}\)
Explanation:
In given equation, \({\dfrac{\alpha z}{k \theta}}\) should be dimensionless \({\therefore \alpha=\dfrac{k \theta}{z} \Rightarrow[\alpha]=\dfrac{\left[M L^{2} T^{-2} K^{-1} \times K\right]}{[L]}}\) \({=\left[M L T^{-2}\right]}\) and \({P=\dfrac{\alpha}{\beta} \Rightarrow[\beta]=\left[\dfrac{\alpha}{P}\right]=\dfrac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]}}\) \({=\left[M^{0} L^{2} T^{0}\right]}\)
PHXI02:UNITS AND MEASUREMENTS
367292
Given that the displacement of an oscillating particle is given by \(y=A \sin (B x+C t+D)\). The dimensional formula for \((A B C D)\) is:
1 \(\left[M^{0} L^{-1} T^{0}\right]\)
2 \(\left[M^{0} L^{0} T^{-1}\right]\)
3 \(\left[M^{0} L^{-1} T^{-1}\right]\)
4 \(\left[M^{0} L^{0} T^{0}\right]\)
Explanation:
Trigonometric function are dimensionless \(D=M^{0} L^{0} T^{0}\) \(C=\dfrac{1}{T}=M^{0} L^{0} T^{-1}\) \(B=\dfrac{1}{x}=M^{0} L^{-1} T^{0}\) \(A=\) Dimension of \(y=M^{0} L^{1} T^{0}\) \([A B C D]=M^{0} L^{0} T^{-1}\).
367289
In relation \(P = \frac{a}{b}{e^{ - \left( {aE/t} \right)}}\) \(P\) represents number, \(E\) is energy and \(t\) is time. The dimension of \(a\) and \(b\) is
As \(P\) is a number then \(a\) and \(b\) have same units \(\frac{{aE}}{t} = {M^o}{L^o}{T^o}\) \(\frac{{a\left( {M{L^2}{T^{ - 2}}} \right)}}{T} = {M^o}{L^o}{T^o}\) \(a = ({M^{ - 1}}{L^{ - 2}}{T^3}) = b\)
PHXI02:UNITS AND MEASUREMENTS
367290
Find the dimensional formula of surface tension \(S\),equation involving the surface tension \(S = \frac{{\rho grh}}{2}\)
367291
In the relation, \({P=\dfrac{\alpha}{\beta} e^{-\dfrac{\alpha}{k \theta}} P}\) is pressure, \({Z}\) is the distance, \({k}\) is Boltzmann constant, and \({\theta}\) is the temperature. The dimensional formula of \({\beta}\) will be
1 \({\left[M^{0} L^{2} T^{0}\right]}\)
2 \({\left[M^{1} L^{2} T^{1}\right]}\)
3 \({\left[M^{1} L^{0} T^{-1}\right]}\)
4 \({\left[M^{1} L^{2} T^{-1}\right]}\)
Explanation:
In given equation, \({\dfrac{\alpha z}{k \theta}}\) should be dimensionless \({\therefore \alpha=\dfrac{k \theta}{z} \Rightarrow[\alpha]=\dfrac{\left[M L^{2} T^{-2} K^{-1} \times K\right]}{[L]}}\) \({=\left[M L T^{-2}\right]}\) and \({P=\dfrac{\alpha}{\beta} \Rightarrow[\beta]=\left[\dfrac{\alpha}{P}\right]=\dfrac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]}}\) \({=\left[M^{0} L^{2} T^{0}\right]}\)
PHXI02:UNITS AND MEASUREMENTS
367292
Given that the displacement of an oscillating particle is given by \(y=A \sin (B x+C t+D)\). The dimensional formula for \((A B C D)\) is:
1 \(\left[M^{0} L^{-1} T^{0}\right]\)
2 \(\left[M^{0} L^{0} T^{-1}\right]\)
3 \(\left[M^{0} L^{-1} T^{-1}\right]\)
4 \(\left[M^{0} L^{0} T^{0}\right]\)
Explanation:
Trigonometric function are dimensionless \(D=M^{0} L^{0} T^{0}\) \(C=\dfrac{1}{T}=M^{0} L^{0} T^{-1}\) \(B=\dfrac{1}{x}=M^{0} L^{-1} T^{0}\) \(A=\) Dimension of \(y=M^{0} L^{1} T^{0}\) \([A B C D]=M^{0} L^{0} T^{-1}\).
367289
In relation \(P = \frac{a}{b}{e^{ - \left( {aE/t} \right)}}\) \(P\) represents number, \(E\) is energy and \(t\) is time. The dimension of \(a\) and \(b\) is
As \(P\) is a number then \(a\) and \(b\) have same units \(\frac{{aE}}{t} = {M^o}{L^o}{T^o}\) \(\frac{{a\left( {M{L^2}{T^{ - 2}}} \right)}}{T} = {M^o}{L^o}{T^o}\) \(a = ({M^{ - 1}}{L^{ - 2}}{T^3}) = b\)
PHXI02:UNITS AND MEASUREMENTS
367290
Find the dimensional formula of surface tension \(S\),equation involving the surface tension \(S = \frac{{\rho grh}}{2}\)
367291
In the relation, \({P=\dfrac{\alpha}{\beta} e^{-\dfrac{\alpha}{k \theta}} P}\) is pressure, \({Z}\) is the distance, \({k}\) is Boltzmann constant, and \({\theta}\) is the temperature. The dimensional formula of \({\beta}\) will be
1 \({\left[M^{0} L^{2} T^{0}\right]}\)
2 \({\left[M^{1} L^{2} T^{1}\right]}\)
3 \({\left[M^{1} L^{0} T^{-1}\right]}\)
4 \({\left[M^{1} L^{2} T^{-1}\right]}\)
Explanation:
In given equation, \({\dfrac{\alpha z}{k \theta}}\) should be dimensionless \({\therefore \alpha=\dfrac{k \theta}{z} \Rightarrow[\alpha]=\dfrac{\left[M L^{2} T^{-2} K^{-1} \times K\right]}{[L]}}\) \({=\left[M L T^{-2}\right]}\) and \({P=\dfrac{\alpha}{\beta} \Rightarrow[\beta]=\left[\dfrac{\alpha}{P}\right]=\dfrac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]}}\) \({=\left[M^{0} L^{2} T^{0}\right]}\)
PHXI02:UNITS AND MEASUREMENTS
367292
Given that the displacement of an oscillating particle is given by \(y=A \sin (B x+C t+D)\). The dimensional formula for \((A B C D)\) is:
1 \(\left[M^{0} L^{-1} T^{0}\right]\)
2 \(\left[M^{0} L^{0} T^{-1}\right]\)
3 \(\left[M^{0} L^{-1} T^{-1}\right]\)
4 \(\left[M^{0} L^{0} T^{0}\right]\)
Explanation:
Trigonometric function are dimensionless \(D=M^{0} L^{0} T^{0}\) \(C=\dfrac{1}{T}=M^{0} L^{0} T^{-1}\) \(B=\dfrac{1}{x}=M^{0} L^{-1} T^{0}\) \(A=\) Dimension of \(y=M^{0} L^{1} T^{0}\) \([A B C D]=M^{0} L^{0} T^{-1}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI02:UNITS AND MEASUREMENTS
367289
In relation \(P = \frac{a}{b}{e^{ - \left( {aE/t} \right)}}\) \(P\) represents number, \(E\) is energy and \(t\) is time. The dimension of \(a\) and \(b\) is
As \(P\) is a number then \(a\) and \(b\) have same units \(\frac{{aE}}{t} = {M^o}{L^o}{T^o}\) \(\frac{{a\left( {M{L^2}{T^{ - 2}}} \right)}}{T} = {M^o}{L^o}{T^o}\) \(a = ({M^{ - 1}}{L^{ - 2}}{T^3}) = b\)
PHXI02:UNITS AND MEASUREMENTS
367290
Find the dimensional formula of surface tension \(S\),equation involving the surface tension \(S = \frac{{\rho grh}}{2}\)
367291
In the relation, \({P=\dfrac{\alpha}{\beta} e^{-\dfrac{\alpha}{k \theta}} P}\) is pressure, \({Z}\) is the distance, \({k}\) is Boltzmann constant, and \({\theta}\) is the temperature. The dimensional formula of \({\beta}\) will be
1 \({\left[M^{0} L^{2} T^{0}\right]}\)
2 \({\left[M^{1} L^{2} T^{1}\right]}\)
3 \({\left[M^{1} L^{0} T^{-1}\right]}\)
4 \({\left[M^{1} L^{2} T^{-1}\right]}\)
Explanation:
In given equation, \({\dfrac{\alpha z}{k \theta}}\) should be dimensionless \({\therefore \alpha=\dfrac{k \theta}{z} \Rightarrow[\alpha]=\dfrac{\left[M L^{2} T^{-2} K^{-1} \times K\right]}{[L]}}\) \({=\left[M L T^{-2}\right]}\) and \({P=\dfrac{\alpha}{\beta} \Rightarrow[\beta]=\left[\dfrac{\alpha}{P}\right]=\dfrac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]}}\) \({=\left[M^{0} L^{2} T^{0}\right]}\)
PHXI02:UNITS AND MEASUREMENTS
367292
Given that the displacement of an oscillating particle is given by \(y=A \sin (B x+C t+D)\). The dimensional formula for \((A B C D)\) is:
1 \(\left[M^{0} L^{-1} T^{0}\right]\)
2 \(\left[M^{0} L^{0} T^{-1}\right]\)
3 \(\left[M^{0} L^{-1} T^{-1}\right]\)
4 \(\left[M^{0} L^{0} T^{0}\right]\)
Explanation:
Trigonometric function are dimensionless \(D=M^{0} L^{0} T^{0}\) \(C=\dfrac{1}{T}=M^{0} L^{0} T^{-1}\) \(B=\dfrac{1}{x}=M^{0} L^{-1} T^{0}\) \(A=\) Dimension of \(y=M^{0} L^{1} T^{0}\) \([A B C D]=M^{0} L^{0} T^{-1}\).