Given least count of Screw Gauge =0.01 mm L.C = No. of circular turn ( pitch )=NP=0.01 mm New pitch =N(1−0.5)P(1+0.75)=NP[0.51.75]=(0.01)3.5=0.035 mm=35×10−3 mm
The least count of a screw guage is 0.01 mm . If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be _____ ×10−3 mm
Held on 24 Jan 2025 · Verified 6 Jul 2026.
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Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
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The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
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