The distance traveled by a freely falling body is given by the equation- y - v0 t -1-2gt2- Using principles of dimensions- check the correctness of the equation-

Check the correctness of the given equationy = v_0 t –(1/2)gt^2By, LHSy=[M^0L^1T^0]=[L]By, RHS⇒v_0 t –(1/2)gt^2=[M^0L^1T^1 1] [M^0L^1T^2]×[M^0L^0T^ 2]0ex0ex⇒v_0 ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The distance ...

Question

The distance traveled by a freely falling body is given by the equation, $y = v_{0} t - (\frac{1}{2}) g t^{2}$ . Using principles of dimensions, check the correctness of the equation.

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Solution

Check the correctness of the given equation
$y = v_{0} t - (\frac{1}{2}) g t^{2}$
By, LHS
$y = [M^{0} L^{1} T^{0}] = [L]$
By, RHS
$\Rightarrow v_{0} t - (\frac{1}{2}) g t^{2} = [M^{0} L^{1} T^{1 - 1}] - [M^{0} L^{1} T^{2}] \times [M^{0} L^{0} T^{- 2}] \Rightarrow v_{0} t - (\frac{1}{2}) g t^{2} = [M^{0} L^{1} T^{0}] - [M^{0} L^{1} T^{2 - 2}] \Rightarrow v_{0} t - (\frac{1}{2}) g t^{2} = [M^{0} L^{1} T^{0}] - [M^{0} L^{1} T^{0}] = [L]$
So, LHS $=$ RHS
Hence, the equation is correct by the principle of dimensions.

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The distance traveled by a freely falling body is given by the equation, y=v0t–12gt2. Using principles of dimensions, check the correctness of the equation.

Check the correctness of the given equationy=v0t–12gt2By, LHSy=M0L1T0=LBy, RHS⇒v0t–12gt2=M0L1T1-1-M0L1T2×M0L0T-2⇒v0t–12gt2=M0L1T0-M0L1T2-2⇒v0t–12gt2=M0L1T0-M0L1T0=LSo, LHS=RHSHence, the equation is correct by the principle of dimensions.

The distance traveled by a freely falling body is given by the equation, $y = v_{0} t - (\frac{1}{2}) g t^{2}$ . Using principles of dimensions, check the correctness of the equation.