If the acceleration due to gravity is $10\,m{s^{ - 2}}$ and the units of length and time are changed in kilometer and hour respectively, the numerical value of the acceleration is
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 SolutionWhich of the following is not represented in correct unit
- 2View SolutionWhich of the following is the most precise device for measuring length
- 3The quantity $X = \frac{{{\varepsilon _0}LV}}{t}$: ${\varepsilon _0}$ is the permittivity of free space, $L$ is length, $V$ is potential difference and $t$ is time. The dimensions of $X$ are same as that ofView Solution
- 4The least count of the main scale of a screw gauge is $1\, mm$. The minimum number of divisions on its circular scale required to measure $5\,\mu m$ diameter of a wire isView Solution
- 5In an experiment, the following observation's were recorded : $L = 2.820\, m, M = 3.00 \,kg, l = 0.087 \,cm$, Diameter $D = 0.041 \,cm$ Taking $g = 9.81$ $m/{s^2}$ using the formula , $Y=\frac{{4MgL}}{{\pi {D^2}l}}$, the maximum permissible error in $Y$ is ......... $\%$View Solution
- 6If $x$ and $a$ stand for distance then for what value of $n$ is given equation dimensionally correct the eq. is $\int {\frac{{dx}}{{\sqrt {{a^2}\, - \,{x^n}} \,}}\, = \,{{\sin }^{ - 1}}\,\frac{x}{a}} $View Solution
- 7Dimension of $\frac{1}{\mu_0 \varepsilon_0}$ should be equal toView Solution
- 8An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment time for $100\, oscillations$ is measured by using a watch of $1\, second$ least count and the value is $90.0\, seconds$. The length $L$ is measured by using a meter scale of least count $1\, mm$ and the value is $20.0\, cm$. The error in the determination of $g$ would be ........... $\%$View Solution
- 9View SolutionWhich one of the following quantities has dimensions different from the remaining three
- 10A massive black hole of mass $m$ and radius $R$ is spinning with angular velocity $\omega$. The power $P$ radiated by it as gravitational waves is given by $P=G c^{-5} m^{x} R^{y} \omega^{z}$, where $c$ and $G$ are speed of light in free space and the universal gravitational constant, respectively. Then,View Solution