160569
A student measured the length of a rod and wrote it as \(3.55 \mathrm{~cm}\). Which instrument did he use to measure it :
1 A meter scale of least cound \(1 \mathrm{~cm}\)
2 A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 5 divisions in \(1 \mathrm{~cm}\)
3 A screw gauge having 100 divisions in the circular scale and pitch as \(1 \mathrm{~mm}\)
4 A screw gauge having 10 divisons in the circular scale and pitch as \(1 \mathrm{~mm}\).
Explanation:
For measurement of \(3.55 \mathrm{~cm}\) least count should be \(0.01 \mathrm{~cm}\) or \(0.05 \mathrm{~cm}\) only. \(\therefore\) Screw gauge of pitch \(5 \mathrm{~mm}\) and circular division \(=10\) \(L C=5 / 10=0.5 \mathrm{~mm}=0.05 \mathrm{~cm}\).
NCERT-I-18
1 RBTS PAPER
160570
Assertion (A): \(\frac{L}{R}\) and \(C R\) both have same dimensions Reason (R) : \(\frac{\mathrm{L}}{\mathrm{R}}\) and CR both have dimension of time.
1 Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\)
2 Both \(A\) and \(R\) are true but \(R\) is not the correct explanation of \(A\).
3 \(A\) is true but \(R\) is false
4 \(A\) is false but \(R\) is true
Explanation:
Unit of quantity \(\left(\frac{L}{R}\right)\) is henry/ohm. As henry \(=\) ohm \(x\) sec hence unit of \(\left[\frac{L}{R}\right]\) is sec i.e. \(\left[\frac{L}{R}\right]=[T]\) Similarly, unit of product CR is farad ohm or, \(\frac{\text { Columb }}{\text { volt }} \times \frac{\text { volt }}{\text { amp }}\) or \(\frac{\text { sex } \times \text { amp }}{a m p}\) or, sec i.e. \([C R]=[T]\) therefore [] and \([C R]\) both have the same dimensions.
NCERT-I-158
1 RBTS PAPER
160571
If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula.
According to the problem, fundamental quantities are momentum ( \(p\) ), area (A) and time ( \(T\) ) and we have to express energy in these fundamental quantities. Let energy \(\mathrm{E}\), \(E \propto p^a A^A T^C \Rightarrow E=k p^a A^A T^C\) where, \(k\) is dimensionless constant of proportionality. Dimensional formula of energy, \([E]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\) and \([\mathrm{p}]=\left[\mathrm{MLT}^{-1}\right]\) \([A]=\left[L^2\right],[T]=[T]\) and \([E]=[K][p]^a[A]^b[T]^c\) Putting all the dimensions, we get \(\begin{aligned} \mathrm{ML}^2 \mathrm{~T}^2 & =\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^2\right]^{\mathrm{b}}[T]^{\mathrm{c}} \\ & =\mathrm{M}^{\mathrm{a}} \mathrm{M}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{\mathrm{a}+\mathrm{c}} \end{aligned}\) According to the principle of homogeneity of dimensions, we get \(\begin{aligned} & a=1 \ldots \ldots(i) \\ & a+2 b=2 \\ & -a+c=-2 . \end{aligned}\) By solving these equations (i), (ii) and (iii), we get \(a=1, b=1 / 2, c=-1\) Dimesnional formula for \(E\) is \(\left[p^1 A^{1 / 2} T^{-1}\right]\)
160569
A student measured the length of a rod and wrote it as \(3.55 \mathrm{~cm}\). Which instrument did he use to measure it :
1 A meter scale of least cound \(1 \mathrm{~cm}\)
2 A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 5 divisions in \(1 \mathrm{~cm}\)
3 A screw gauge having 100 divisions in the circular scale and pitch as \(1 \mathrm{~mm}\)
4 A screw gauge having 10 divisons in the circular scale and pitch as \(1 \mathrm{~mm}\).
Explanation:
For measurement of \(3.55 \mathrm{~cm}\) least count should be \(0.01 \mathrm{~cm}\) or \(0.05 \mathrm{~cm}\) only. \(\therefore\) Screw gauge of pitch \(5 \mathrm{~mm}\) and circular division \(=10\) \(L C=5 / 10=0.5 \mathrm{~mm}=0.05 \mathrm{~cm}\).
NCERT-I-18
1 RBTS PAPER
160570
Assertion (A): \(\frac{L}{R}\) and \(C R\) both have same dimensions Reason (R) : \(\frac{\mathrm{L}}{\mathrm{R}}\) and CR both have dimension of time.
1 Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\)
2 Both \(A\) and \(R\) are true but \(R\) is not the correct explanation of \(A\).
3 \(A\) is true but \(R\) is false
4 \(A\) is false but \(R\) is true
Explanation:
Unit of quantity \(\left(\frac{L}{R}\right)\) is henry/ohm. As henry \(=\) ohm \(x\) sec hence unit of \(\left[\frac{L}{R}\right]\) is sec i.e. \(\left[\frac{L}{R}\right]=[T]\) Similarly, unit of product CR is farad ohm or, \(\frac{\text { Columb }}{\text { volt }} \times \frac{\text { volt }}{\text { amp }}\) or \(\frac{\text { sex } \times \text { amp }}{a m p}\) or, sec i.e. \([C R]=[T]\) therefore [] and \([C R]\) both have the same dimensions.
NCERT-I-158
1 RBTS PAPER
160571
If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula.
According to the problem, fundamental quantities are momentum ( \(p\) ), area (A) and time ( \(T\) ) and we have to express energy in these fundamental quantities. Let energy \(\mathrm{E}\), \(E \propto p^a A^A T^C \Rightarrow E=k p^a A^A T^C\) where, \(k\) is dimensionless constant of proportionality. Dimensional formula of energy, \([E]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\) and \([\mathrm{p}]=\left[\mathrm{MLT}^{-1}\right]\) \([A]=\left[L^2\right],[T]=[T]\) and \([E]=[K][p]^a[A]^b[T]^c\) Putting all the dimensions, we get \(\begin{aligned} \mathrm{ML}^2 \mathrm{~T}^2 & =\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^2\right]^{\mathrm{b}}[T]^{\mathrm{c}} \\ & =\mathrm{M}^{\mathrm{a}} \mathrm{M}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{\mathrm{a}+\mathrm{c}} \end{aligned}\) According to the principle of homogeneity of dimensions, we get \(\begin{aligned} & a=1 \ldots \ldots(i) \\ & a+2 b=2 \\ & -a+c=-2 . \end{aligned}\) By solving these equations (i), (ii) and (iii), we get \(a=1, b=1 / 2, c=-1\) Dimesnional formula for \(E\) is \(\left[p^1 A^{1 / 2} T^{-1}\right]\)
160569
A student measured the length of a rod and wrote it as \(3.55 \mathrm{~cm}\). Which instrument did he use to measure it :
1 A meter scale of least cound \(1 \mathrm{~cm}\)
2 A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 5 divisions in \(1 \mathrm{~cm}\)
3 A screw gauge having 100 divisions in the circular scale and pitch as \(1 \mathrm{~mm}\)
4 A screw gauge having 10 divisons in the circular scale and pitch as \(1 \mathrm{~mm}\).
Explanation:
For measurement of \(3.55 \mathrm{~cm}\) least count should be \(0.01 \mathrm{~cm}\) or \(0.05 \mathrm{~cm}\) only. \(\therefore\) Screw gauge of pitch \(5 \mathrm{~mm}\) and circular division \(=10\) \(L C=5 / 10=0.5 \mathrm{~mm}=0.05 \mathrm{~cm}\).
NCERT-I-18
1 RBTS PAPER
160570
Assertion (A): \(\frac{L}{R}\) and \(C R\) both have same dimensions Reason (R) : \(\frac{\mathrm{L}}{\mathrm{R}}\) and CR both have dimension of time.
1 Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\)
2 Both \(A\) and \(R\) are true but \(R\) is not the correct explanation of \(A\).
3 \(A\) is true but \(R\) is false
4 \(A\) is false but \(R\) is true
Explanation:
Unit of quantity \(\left(\frac{L}{R}\right)\) is henry/ohm. As henry \(=\) ohm \(x\) sec hence unit of \(\left[\frac{L}{R}\right]\) is sec i.e. \(\left[\frac{L}{R}\right]=[T]\) Similarly, unit of product CR is farad ohm or, \(\frac{\text { Columb }}{\text { volt }} \times \frac{\text { volt }}{\text { amp }}\) or \(\frac{\text { sex } \times \text { amp }}{a m p}\) or, sec i.e. \([C R]=[T]\) therefore [] and \([C R]\) both have the same dimensions.
NCERT-I-158
1 RBTS PAPER
160571
If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula.
According to the problem, fundamental quantities are momentum ( \(p\) ), area (A) and time ( \(T\) ) and we have to express energy in these fundamental quantities. Let energy \(\mathrm{E}\), \(E \propto p^a A^A T^C \Rightarrow E=k p^a A^A T^C\) where, \(k\) is dimensionless constant of proportionality. Dimensional formula of energy, \([E]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\) and \([\mathrm{p}]=\left[\mathrm{MLT}^{-1}\right]\) \([A]=\left[L^2\right],[T]=[T]\) and \([E]=[K][p]^a[A]^b[T]^c\) Putting all the dimensions, we get \(\begin{aligned} \mathrm{ML}^2 \mathrm{~T}^2 & =\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^2\right]^{\mathrm{b}}[T]^{\mathrm{c}} \\ & =\mathrm{M}^{\mathrm{a}} \mathrm{M}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{\mathrm{a}+\mathrm{c}} \end{aligned}\) According to the principle of homogeneity of dimensions, we get \(\begin{aligned} & a=1 \ldots \ldots(i) \\ & a+2 b=2 \\ & -a+c=-2 . \end{aligned}\) By solving these equations (i), (ii) and (iii), we get \(a=1, b=1 / 2, c=-1\) Dimesnional formula for \(E\) is \(\left[p^1 A^{1 / 2} T^{-1}\right]\)
160569
A student measured the length of a rod and wrote it as \(3.55 \mathrm{~cm}\). Which instrument did he use to measure it :
1 A meter scale of least cound \(1 \mathrm{~cm}\)
2 A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 5 divisions in \(1 \mathrm{~cm}\)
3 A screw gauge having 100 divisions in the circular scale and pitch as \(1 \mathrm{~mm}\)
4 A screw gauge having 10 divisons in the circular scale and pitch as \(1 \mathrm{~mm}\).
Explanation:
For measurement of \(3.55 \mathrm{~cm}\) least count should be \(0.01 \mathrm{~cm}\) or \(0.05 \mathrm{~cm}\) only. \(\therefore\) Screw gauge of pitch \(5 \mathrm{~mm}\) and circular division \(=10\) \(L C=5 / 10=0.5 \mathrm{~mm}=0.05 \mathrm{~cm}\).
NCERT-I-18
1 RBTS PAPER
160570
Assertion (A): \(\frac{L}{R}\) and \(C R\) both have same dimensions Reason (R) : \(\frac{\mathrm{L}}{\mathrm{R}}\) and CR both have dimension of time.
1 Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\)
2 Both \(A\) and \(R\) are true but \(R\) is not the correct explanation of \(A\).
3 \(A\) is true but \(R\) is false
4 \(A\) is false but \(R\) is true
Explanation:
Unit of quantity \(\left(\frac{L}{R}\right)\) is henry/ohm. As henry \(=\) ohm \(x\) sec hence unit of \(\left[\frac{L}{R}\right]\) is sec i.e. \(\left[\frac{L}{R}\right]=[T]\) Similarly, unit of product CR is farad ohm or, \(\frac{\text { Columb }}{\text { volt }} \times \frac{\text { volt }}{\text { amp }}\) or \(\frac{\text { sex } \times \text { amp }}{a m p}\) or, sec i.e. \([C R]=[T]\) therefore [] and \([C R]\) both have the same dimensions.
NCERT-I-158
1 RBTS PAPER
160571
If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula.
According to the problem, fundamental quantities are momentum ( \(p\) ), area (A) and time ( \(T\) ) and we have to express energy in these fundamental quantities. Let energy \(\mathrm{E}\), \(E \propto p^a A^A T^C \Rightarrow E=k p^a A^A T^C\) where, \(k\) is dimensionless constant of proportionality. Dimensional formula of energy, \([E]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\) and \([\mathrm{p}]=\left[\mathrm{MLT}^{-1}\right]\) \([A]=\left[L^2\right],[T]=[T]\) and \([E]=[K][p]^a[A]^b[T]^c\) Putting all the dimensions, we get \(\begin{aligned} \mathrm{ML}^2 \mathrm{~T}^2 & =\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^2\right]^{\mathrm{b}}[T]^{\mathrm{c}} \\ & =\mathrm{M}^{\mathrm{a}} \mathrm{M}^{\mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{\mathrm{a}+\mathrm{c}} \end{aligned}\) According to the principle of homogeneity of dimensions, we get \(\begin{aligned} & a=1 \ldots \ldots(i) \\ & a+2 b=2 \\ & -a+c=-2 . \end{aligned}\) By solving these equations (i), (ii) and (iii), we get \(a=1, b=1 / 2, c=-1\) Dimesnional formula for \(E\) is \(\left[p^1 A^{1 / 2} T^{-1}\right]\)