367286
Consider two physical quantities \(A\) and \(B\) related to each other as \(E=\dfrac{B-x^{2}}{A t}\) where \(E,x\) and \(t\) have dimensions of energy, length and time respectively.The dimension of \(AB\) is
1 \(L^{0} M^{-1} T^{1}\)
2 \(L^{-2} M^{1} T^{0}\)
3 \(L^{2} M^{-1} T^{1}\)
4 \(L^{-2} M^{-1} T^{1}\)
Explanation:
Given, \(E=\dfrac{B-x^{2}}{A t}\) Using principle of homogeneity, we can say \(B\) and \(x^{2}\) should have same dimension i.e., \([B]=\left[L^{2}\right]\) Also, \([A]=\dfrac{\left[B-x^{2}\right]}{[E t]}=\dfrac{\left[L^{2}\right]}{\left[M^{1} L^{2} T^{-2}\right]\left[T^{1}\right]}\) \(A=\left[M^{-1} T\right]\) Now, \([A B]=\left[M^{-1} T\right]\left[L^{2}\right]\) \(=\left[M^{-1} L^{2} T^{1}\right]\)
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367287
Distance \({Z}\) travelled by a particle is defined by \({Z=\alpha+\beta t+\gamma t^{2}}\). Dimensions of \({\gamma}\) are
1 \({\left[{LT}^{-1}\right]}\)
2 \({\left[{L}^{-1} {~T}\right]}\)
3 \({\left[{LT}^{-2}\right]}\)
4 \({\left[{LT}^{2}\right]}\)
Explanation:
\({ Z=\alpha+\beta t+\gamma t^{2}}\) \([Z]=[\alpha]=[\beta t]=\left[\gamma t^{2}\right]\) \({\therefore \quad[\gamma]=[Z][t]^{-2}=\left[{LT}^{-2}\right]}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367288
The dimension of physical quantity \(X\) in the equation force \(=\dfrac{X}{\sqrt{\text { Density }}}\) is given by:
367286
Consider two physical quantities \(A\) and \(B\) related to each other as \(E=\dfrac{B-x^{2}}{A t}\) where \(E,x\) and \(t\) have dimensions of energy, length and time respectively.The dimension of \(AB\) is
1 \(L^{0} M^{-1} T^{1}\)
2 \(L^{-2} M^{1} T^{0}\)
3 \(L^{2} M^{-1} T^{1}\)
4 \(L^{-2} M^{-1} T^{1}\)
Explanation:
Given, \(E=\dfrac{B-x^{2}}{A t}\) Using principle of homogeneity, we can say \(B\) and \(x^{2}\) should have same dimension i.e., \([B]=\left[L^{2}\right]\) Also, \([A]=\dfrac{\left[B-x^{2}\right]}{[E t]}=\dfrac{\left[L^{2}\right]}{\left[M^{1} L^{2} T^{-2}\right]\left[T^{1}\right]}\) \(A=\left[M^{-1} T\right]\) Now, \([A B]=\left[M^{-1} T\right]\left[L^{2}\right]\) \(=\left[M^{-1} L^{2} T^{1}\right]\)
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367287
Distance \({Z}\) travelled by a particle is defined by \({Z=\alpha+\beta t+\gamma t^{2}}\). Dimensions of \({\gamma}\) are
1 \({\left[{LT}^{-1}\right]}\)
2 \({\left[{L}^{-1} {~T}\right]}\)
3 \({\left[{LT}^{-2}\right]}\)
4 \({\left[{LT}^{2}\right]}\)
Explanation:
\({ Z=\alpha+\beta t+\gamma t^{2}}\) \([Z]=[\alpha]=[\beta t]=\left[\gamma t^{2}\right]\) \({\therefore \quad[\gamma]=[Z][t]^{-2}=\left[{LT}^{-2}\right]}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367288
The dimension of physical quantity \(X\) in the equation force \(=\dfrac{X}{\sqrt{\text { Density }}}\) is given by:
367286
Consider two physical quantities \(A\) and \(B\) related to each other as \(E=\dfrac{B-x^{2}}{A t}\) where \(E,x\) and \(t\) have dimensions of energy, length and time respectively.The dimension of \(AB\) is
1 \(L^{0} M^{-1} T^{1}\)
2 \(L^{-2} M^{1} T^{0}\)
3 \(L^{2} M^{-1} T^{1}\)
4 \(L^{-2} M^{-1} T^{1}\)
Explanation:
Given, \(E=\dfrac{B-x^{2}}{A t}\) Using principle of homogeneity, we can say \(B\) and \(x^{2}\) should have same dimension i.e., \([B]=\left[L^{2}\right]\) Also, \([A]=\dfrac{\left[B-x^{2}\right]}{[E t]}=\dfrac{\left[L^{2}\right]}{\left[M^{1} L^{2} T^{-2}\right]\left[T^{1}\right]}\) \(A=\left[M^{-1} T\right]\) Now, \([A B]=\left[M^{-1} T\right]\left[L^{2}\right]\) \(=\left[M^{-1} L^{2} T^{1}\right]\)
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367287
Distance \({Z}\) travelled by a particle is defined by \({Z=\alpha+\beta t+\gamma t^{2}}\). Dimensions of \({\gamma}\) are
1 \({\left[{LT}^{-1}\right]}\)
2 \({\left[{L}^{-1} {~T}\right]}\)
3 \({\left[{LT}^{-2}\right]}\)
4 \({\left[{LT}^{2}\right]}\)
Explanation:
\({ Z=\alpha+\beta t+\gamma t^{2}}\) \([Z]=[\alpha]=[\beta t]=\left[\gamma t^{2}\right]\) \({\therefore \quad[\gamma]=[Z][t]^{-2}=\left[{LT}^{-2}\right]}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367288
The dimension of physical quantity \(X\) in the equation force \(=\dfrac{X}{\sqrt{\text { Density }}}\) is given by:
367286
Consider two physical quantities \(A\) and \(B\) related to each other as \(E=\dfrac{B-x^{2}}{A t}\) where \(E,x\) and \(t\) have dimensions of energy, length and time respectively.The dimension of \(AB\) is
1 \(L^{0} M^{-1} T^{1}\)
2 \(L^{-2} M^{1} T^{0}\)
3 \(L^{2} M^{-1} T^{1}\)
4 \(L^{-2} M^{-1} T^{1}\)
Explanation:
Given, \(E=\dfrac{B-x^{2}}{A t}\) Using principle of homogeneity, we can say \(B\) and \(x^{2}\) should have same dimension i.e., \([B]=\left[L^{2}\right]\) Also, \([A]=\dfrac{\left[B-x^{2}\right]}{[E t]}=\dfrac{\left[L^{2}\right]}{\left[M^{1} L^{2} T^{-2}\right]\left[T^{1}\right]}\) \(A=\left[M^{-1} T\right]\) Now, \([A B]=\left[M^{-1} T\right]\left[L^{2}\right]\) \(=\left[M^{-1} L^{2} T^{1}\right]\)
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367287
Distance \({Z}\) travelled by a particle is defined by \({Z=\alpha+\beta t+\gamma t^{2}}\). Dimensions of \({\gamma}\) are
1 \({\left[{LT}^{-1}\right]}\)
2 \({\left[{L}^{-1} {~T}\right]}\)
3 \({\left[{LT}^{-2}\right]}\)
4 \({\left[{LT}^{2}\right]}\)
Explanation:
\({ Z=\alpha+\beta t+\gamma t^{2}}\) \([Z]=[\alpha]=[\beta t]=\left[\gamma t^{2}\right]\) \({\therefore \quad[\gamma]=[Z][t]^{-2}=\left[{LT}^{-2}\right]}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367288
The dimension of physical quantity \(X\) in the equation force \(=\dfrac{X}{\sqrt{\text { Density }}}\) is given by: