The values of kinetic energy $$K$$ and potential energy $$U$$ are measured as follows :
$$K = 100.0 \pm 2.0\,J,U = 200.0 \pm 1.0\,J.$$ Then the
percentage error in the measurement of mechanical energy is -
A.
$$2.5\% $$
B.
$$1\% $$
C.
$$0.5\% $$
D.
$$1.5\% $$
Answer :
$$1\% $$
Solution :
$$\eqalign{
& {M_E} = K + U = 300 \pm 3J = E \pm \Delta E \cr
& \therefore \frac{{\Delta E}}{E} \times 100 = 1\% \cr} $$
Releted MCQ Question on Basic Physics >> Unit and Measurement
Releted Question 1
The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)
A quantity $$X$$ is given by $${\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}}$$ where $${ \in _0}$$ is the permittivity of the free space, $$L$$ is a length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of-
Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-