A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90\;s$ ,$91\;s $, $95\;s$ and $92\;s$. If the minimum division in the measuring clock is $1\;s$, then the reported mean time should be
JEE MAIN 2016, Medium
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionDimensional formula of resistivity is
- 2$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-View Solution
- 3View SolutionThe dimensional formula of angular impulse is:
- 4View SolutionWhich of the following does not have dimensions of force?
- 5The percentage errors in quantities $P, Q, R$ and $S$ are $0.5\%,\,1\%,\,3\%$ and $1 .5\%$ respectively in the measurement of a physical quantity $A\, = \,\frac{{{P^3}{Q^2}}}{{\sqrt {R}\,S }}$ . the maximum percentage error in the value of $A$ will be $........... \%$View Solution
- 6View SolutionDimension of electric current is
- 7Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.View Solution
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
- 8$Joule-second$ is the unit ofView Solution
- 9If $P$ represents radiation pressure, $c$ represents speed of light and $Q$ represents radiation energy striking a unit area per second, then non-zero integers $x,\,y$ and $z$ such that ${P^x}{Q^y}{c^z}$ is dimensionless, areView Solution
- 10Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers, $5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:View Solution
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.