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1.Units, Dimensions and Measurement
hard
If the present units of length. time and mass $(m, s, k g)$ are changed to $100\; m, 100\; s$. $\frac{1}{10} \;k g$ then
A The new unit of velocity is increased $10$ times
BThe new unit of force is decreased $\frac{1}{1000}$ times
CThe new unit of energy is increased $10$ times
DThe new unit of pressure is increased $1000$ times
Solution
velocity $= m/sec ;$ in new system $= \frac{{100m}}{{100\sec }} = \frac{m}{{\sec }}$ (same)
force $ = \frac{{kg \times m}}{{{{\sec }^2}}}$; in new system $ = \frac{1}{{10}}kg \times \frac{{100m}}{{100\sec \, \times \,100\sec }}$ $ = \frac{1}{{1000}}\,\frac{{kg \times m}}{{{{\sec }^2}}}$
energy $ = \frac{{kg \times {m^2}}}{{{{\sec }^2}}}$ ; in new system $ = \frac{1}{{10}}kg \times \frac{{100m \times 100m}}{{100\sec \, \times \,100\sec }}$ $ = \frac{1}{{10}}\frac{{kg \times {m^{\rm{2}}}}}{{se{c^2}}}$
pressure $ = \frac{{kg}}{{m \times {{\sec }^2}}}$; in new system $ = \frac{1}{{10}}kg \times \frac{1}{{100}}m \times \frac{1}{{100\,sec \times 100\,sec}}$ $ = {10^{ – 7}}\frac{{kg}}{{m \times se{c^2}}}$
force $ = \frac{{kg \times m}}{{{{\sec }^2}}}$; in new system $ = \frac{1}{{10}}kg \times \frac{{100m}}{{100\sec \, \times \,100\sec }}$ $ = \frac{1}{{1000}}\,\frac{{kg \times m}}{{{{\sec }^2}}}$
energy $ = \frac{{kg \times {m^2}}}{{{{\sec }^2}}}$ ; in new system $ = \frac{1}{{10}}kg \times \frac{{100m \times 100m}}{{100\sec \, \times \,100\sec }}$ $ = \frac{1}{{10}}\frac{{kg \times {m^{\rm{2}}}}}{{se{c^2}}}$
pressure $ = \frac{{kg}}{{m \times {{\sec }^2}}}$; in new system $ = \frac{1}{{10}}kg \times \frac{1}{{100}}m \times \frac{1}{{100\,sec \times 100\,sec}}$ $ = {10^{ – 7}}\frac{{kg}}{{m \times se{c^2}}}$
Standard 11
Physics
