If force (F). velocity (V) and time (T) are taken asfundamental units,

1.	If force (F). velocity (V) and time (T) are taken asfundamental units, then the dimensions of massare[AIPMT-2014](1) [F V T-](2) [F VT(3) [FV-T'(4) [F V-T]
Answer» $\maltese \: \underline{\sf AnsWer :} \: \maltese \\$ $\dashrightarrow\:\:\sf m \propto [F]^{x} \: [V]^{y} \: [T]^{z} \\$ By adding 'k' as a proportionality constant we get : $\dashrightarrow\:\:\sf m = k \: [F]^{x} \: [V]^{y} \: [T]^{z} \\$ $\\\bullet\textsf{ Dimensions of mass (m) =\textbf{[M$^ \text1$L$^ \text0$T$^ \text0$]}} \\$ $\bullet\textsf{ Dimensions of force (F) = [M] [LT$^{ \text{-2}}$] = \textbf{[MLT$^{ \text{-2}}$]}} \\$ $\bullet\textsf{ Dimensions of velocity (V) = $\dfrac{ \text{[L]}}{ \text{[T]}}$ = \textbf{[M$^\text0$LT$^{ \text{-1}}$]}} \\$ $\bullet\textsf{ Dimensions of Time Period (T) =\textbf{[M$^ \text0$L$^ \text0$T$^ \text1$]}}\\ \\$ $\dashrightarrow\:\:\sf [M^1 L^0 T^0] = [M^1 L^1 T^{-2}]^{x} \: [M^0 L^1 T^{-1}]^{y} \: [M^0 L^0 T^1]^{z} \\$ $\dashrightarrow\:\:\sf [M^1 L^0 T^0] = [M]^{x} \: [ L]^{x + y} \: [T]^{ - 2x - y + z} \\$ Now, on COMPARING the POWERS of LHS and RHS we get : $\longrightarrow\:\:\sf x = 1 \qquad \qquad...(i)\\$ $\longrightarrow\:\:\sf x + y = 0 \\$ From equation (i) Substituting x = 1 in above equation we get : $\longrightarrow\:\:\sf 1 + y = 0 \\$ $\longrightarrow\:\:\sf y = - 1 \qquad \qquad.... (ii)$ $\longrightarrow\:\:\sf - 2x - y + z = 0 \\$ From equation (i) and (ii) Substituting the VALUE of x and y in above equation: $\longrightarrow\:\:\sf - 2(1) - ( -1 ) + z = 0 \\$ $\longrightarrow\:\:\sf - 2 + 1 + z = 0 \\$ $\longrightarrow\:\:\sf - 2 + 1 = - z\\$ $\longrightarrow\:\:\sf - 1 = - z\\$ $\longrightarrow\:\:\sf z = 1\\$ Hence, our FINAL answer will be : $\dashrightarrow\:\: \underline{ \boxed{\frak{m = [F]^{1} \: [V]^{-1} \: [T]^{1}}}} \\$ Hence,the OPTION (4) is correct answer.

If force (F). velocity (V) and time (T) are taken asfundamental units, then the dimensions of massare[AIPMT-2014](1) [F V T-](2) [F VT(3) [FV-T'(4) [F V-T]

Answer»

$\maltese \: \underline{\sf AnsWer :} \: \maltese \\$

$\dashrightarrow\:\:\sf m \propto [F]^{x} \: [V]^{y} \: [T]^{z} \\$

By adding 'k' as a proportionality constant we get :

$\dashrightarrow\:\:\sf m = k \: [F]^{x} \: [V]^{y} \: [T]^{z} \\$

$\\\bullet\textsf{ Dimensions of mass (m) =\textbf{[M$^ \text1$L$^ \text0$T$^ \text0$]}} \\$

$\bullet\textsf{ Dimensions of force (F) = [M] [LT$^{ \text{-2}}$] = \textbf{[MLT$^{ \text{-2}}$]}} \\$

$\bullet\textsf{ Dimensions of velocity (V) = $\dfrac{ \text{[L]}}{ \text{[T]}}$ = \textbf{[M$^\text0$LT$^{ \text{-1}}$]}} \\$

$\bullet\textsf{ Dimensions of Time Period (T) =\textbf{[M$^ \text0$L$^ \text0$T$^ \text1$]}}\\ \\$

$\dashrightarrow\:\:\sf [M^1 L^0 T^0] = [M^1 L^1 T^{-2}]^{x} \: [M^0 L^1 T^{-1}]^{y} \: [M^0 L^0 T^1]^{z} \\$

$\dashrightarrow\:\:\sf [M^1 L^0 T^0] = [M]^{x} \: [ L]^{x + y} \: [T]^{ - 2x - y + z} \\$

Now, on COMPARING the POWERS of LHS and RHS we get :

$\longrightarrow\:\:\sf x = 1 \qquad \qquad...(i)\\$

$\longrightarrow\:\:\sf x + y = 0 \\$

From equation (i) Substituting x = 1 in above equation we get :

$\longrightarrow\:\:\sf 1 + y = 0 \\$

$\longrightarrow\:\:\sf y = - 1 \qquad \qquad.... (ii)$

$\longrightarrow\:\:\sf - 2x - y + z = 0 \\$

From equation (i) and (ii) Substituting the VALUE of x and y in above equation:

$\longrightarrow\:\:\sf - 2(1) - ( -1 ) + z = 0 \\$

$\longrightarrow\:\:\sf - 2 + 1 + z = 0 \\$

$\longrightarrow\:\:\sf - 2 + 1 = - z\\$

$\longrightarrow\:\:\sf - 1 = - z\\$

$\longrightarrow\:\:\sf z = 1\\$

Hence, our FINAL answer will be :

$\dashrightarrow\:\: \underline{ \boxed{\frak{m = [F]^{1} \: [V]^{-1} \: [T]^{1}}}} \\$

Hence,the OPTION (4) is correct answer.

If force (F). velocity (V) and time (T) are taken asfundamental units, then the dimensions of massare[AIPMT-2014](1) [F V T-](2) [F VT(3) [FV-T'(4) [F V-T]

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If force (F). velocity (V) and time (T) are taken asfundamental units, then the dimensions of massare[AIPMT-2014](1) [F V T-](2) [F VT(3) [FV-T'(4) [F V-T]​

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If force (F). velocity (V) and time (T) are taken asfundamental units, then the dimensions of massare[AIPMT-2014](1) [F V T-](2) [F VT(3) [FV-T'(4) [F V-T]