367320
In the relation: \(y=a \sin (\omega t-k x)\), the dimensional formula for \(k\) is:
1 \(\left[M^{0} L T\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L T^{-1}\right]\)
4 \(\left[M^{0} L^{-1} T^{-1}\right]\)
Explanation:
Here; \(k x\) is dimensionless. Hence, \([k]=\left[\dfrac{2 \pi}{\lambda}\right]=\left[M^{0} L^{-1} T^{0}\right]\) Since \((\omega t-k x)\) is dimensionless.
PHXI02:UNITS AND MEASUREMENTS
367321
In the equation \(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T, X\) is pressure, \(Y\) is volume, \(R\) is universal gas constant and \(T\) is temperature. The physical quantity equivalent to the ratio \(\dfrac{a}{b}\) is
1 impulse
2 coefficient of viscosity
3 energy
4 pressure gradient
Explanation:
\(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T\). As, \(X\) is pressure then its dimensions are\(\left[M L^{-1} T^{-2}\right]\) then \(\dfrac{a}{Y^{2}}\) should also have the same units, using the principle of homogeneity. \(X=\dfrac{a}{Y^{2}}\) \(\left[M L^{-1} T^{-2}\right]=\dfrac{a}{\left[L^{6}\right]}\) \(a=\left[M L^{5} T^{-2}\right]\) Same for \(Y=b\). \(\left[L^{3}\right]=b\) The ratio, \(\dfrac{a}{b}=\dfrac{\left[M L^{5} T^{-2}\right]}{\left[L^{3}\right]}=\left[M L^{2} T^{-2}\right]\). \(\left[M L^{2} T^{-2}\right]\) is the dimension of energy.
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367322
Assertion : A dimensionally wrong or inconsistent equation must be wrong. Reason : A dimensionally consistent equation is an exact or a correct equation.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A dimensionally consistent equation need not be correct equation, but a dimensionally wrong or inconsistent equation must be wrong.So, option (3) is correct.
367320
In the relation: \(y=a \sin (\omega t-k x)\), the dimensional formula for \(k\) is:
1 \(\left[M^{0} L T\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L T^{-1}\right]\)
4 \(\left[M^{0} L^{-1} T^{-1}\right]\)
Explanation:
Here; \(k x\) is dimensionless. Hence, \([k]=\left[\dfrac{2 \pi}{\lambda}\right]=\left[M^{0} L^{-1} T^{0}\right]\) Since \((\omega t-k x)\) is dimensionless.
PHXI02:UNITS AND MEASUREMENTS
367321
In the equation \(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T, X\) is pressure, \(Y\) is volume, \(R\) is universal gas constant and \(T\) is temperature. The physical quantity equivalent to the ratio \(\dfrac{a}{b}\) is
1 impulse
2 coefficient of viscosity
3 energy
4 pressure gradient
Explanation:
\(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T\). As, \(X\) is pressure then its dimensions are\(\left[M L^{-1} T^{-2}\right]\) then \(\dfrac{a}{Y^{2}}\) should also have the same units, using the principle of homogeneity. \(X=\dfrac{a}{Y^{2}}\) \(\left[M L^{-1} T^{-2}\right]=\dfrac{a}{\left[L^{6}\right]}\) \(a=\left[M L^{5} T^{-2}\right]\) Same for \(Y=b\). \(\left[L^{3}\right]=b\) The ratio, \(\dfrac{a}{b}=\dfrac{\left[M L^{5} T^{-2}\right]}{\left[L^{3}\right]}=\left[M L^{2} T^{-2}\right]\). \(\left[M L^{2} T^{-2}\right]\) is the dimension of energy.
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367322
Assertion : A dimensionally wrong or inconsistent equation must be wrong. Reason : A dimensionally consistent equation is an exact or a correct equation.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A dimensionally consistent equation need not be correct equation, but a dimensionally wrong or inconsistent equation must be wrong.So, option (3) is correct.
367320
In the relation: \(y=a \sin (\omega t-k x)\), the dimensional formula for \(k\) is:
1 \(\left[M^{0} L T\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L T^{-1}\right]\)
4 \(\left[M^{0} L^{-1} T^{-1}\right]\)
Explanation:
Here; \(k x\) is dimensionless. Hence, \([k]=\left[\dfrac{2 \pi}{\lambda}\right]=\left[M^{0} L^{-1} T^{0}\right]\) Since \((\omega t-k x)\) is dimensionless.
PHXI02:UNITS AND MEASUREMENTS
367321
In the equation \(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T, X\) is pressure, \(Y\) is volume, \(R\) is universal gas constant and \(T\) is temperature. The physical quantity equivalent to the ratio \(\dfrac{a}{b}\) is
1 impulse
2 coefficient of viscosity
3 energy
4 pressure gradient
Explanation:
\(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T\). As, \(X\) is pressure then its dimensions are\(\left[M L^{-1} T^{-2}\right]\) then \(\dfrac{a}{Y^{2}}\) should also have the same units, using the principle of homogeneity. \(X=\dfrac{a}{Y^{2}}\) \(\left[M L^{-1} T^{-2}\right]=\dfrac{a}{\left[L^{6}\right]}\) \(a=\left[M L^{5} T^{-2}\right]\) Same for \(Y=b\). \(\left[L^{3}\right]=b\) The ratio, \(\dfrac{a}{b}=\dfrac{\left[M L^{5} T^{-2}\right]}{\left[L^{3}\right]}=\left[M L^{2} T^{-2}\right]\). \(\left[M L^{2} T^{-2}\right]\) is the dimension of energy.
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367322
Assertion : A dimensionally wrong or inconsistent equation must be wrong. Reason : A dimensionally consistent equation is an exact or a correct equation.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A dimensionally consistent equation need not be correct equation, but a dimensionally wrong or inconsistent equation must be wrong.So, option (3) is correct.
367320
In the relation: \(y=a \sin (\omega t-k x)\), the dimensional formula for \(k\) is:
1 \(\left[M^{0} L T\right]\)
2 \(\left[M^{0} L^{-1} T^{0}\right]\)
3 \(\left[M^{0} L T^{-1}\right]\)
4 \(\left[M^{0} L^{-1} T^{-1}\right]\)
Explanation:
Here; \(k x\) is dimensionless. Hence, \([k]=\left[\dfrac{2 \pi}{\lambda}\right]=\left[M^{0} L^{-1} T^{0}\right]\) Since \((\omega t-k x)\) is dimensionless.
PHXI02:UNITS AND MEASUREMENTS
367321
In the equation \(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T, X\) is pressure, \(Y\) is volume, \(R\) is universal gas constant and \(T\) is temperature. The physical quantity equivalent to the ratio \(\dfrac{a}{b}\) is
1 impulse
2 coefficient of viscosity
3 energy
4 pressure gradient
Explanation:
\(\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=R T\). As, \(X\) is pressure then its dimensions are\(\left[M L^{-1} T^{-2}\right]\) then \(\dfrac{a}{Y^{2}}\) should also have the same units, using the principle of homogeneity. \(X=\dfrac{a}{Y^{2}}\) \(\left[M L^{-1} T^{-2}\right]=\dfrac{a}{\left[L^{6}\right]}\) \(a=\left[M L^{5} T^{-2}\right]\) Same for \(Y=b\). \(\left[L^{3}\right]=b\) The ratio, \(\dfrac{a}{b}=\dfrac{\left[M L^{5} T^{-2}\right]}{\left[L^{3}\right]}=\left[M L^{2} T^{-2}\right]\). \(\left[M L^{2} T^{-2}\right]\) is the dimension of energy.
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367322
Assertion : A dimensionally wrong or inconsistent equation must be wrong. Reason : A dimensionally consistent equation is an exact or a correct equation.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A dimensionally consistent equation need not be correct equation, but a dimensionally wrong or inconsistent equation must be wrong.So, option (3) is correct.
