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See all →- Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero o… Answer: 3. 4.98 cm Key idea: reading = MSR + (VSR × least count) - zero error. 10 VSD = 9 MSD, so the least count =1 MSD -1 VSD =0.1-0.09=0.... 1
- In a vernier calipers, (N+1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is : Answer: 3. 1/100(N+1) Key idea: Vernier constant = 1 MSD - 1 VSD. N main scale divisions equal N+1 vernier divisions, so 1 VSD=N/N+1 MSD . V... 1
- A screw gauge gives the following readings when used to measure the diameter of a wireMain scale reading : 0 mmCircular scale reading : 52 divisionsGiven that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the… Answer: 4. 0.052 cm Key idea: diameter = main scale reading + circular reading × least count. Least count =1\ mm/100=0.01 mm. Diameter =0+52... 1
- A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.The pitch of the screw gauge is : Answer: 3. 0.5 mm Least count =pitch/number of circular divisions . pitch=0.01\ mm×50=0.5\ mm Answer: option 3. 1
- A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the… Answer: 3. 339 m/s In a resonance tube, two successive resonances are half a wavelength apart. dfracλ2=73-20=53 cm, so λ=106 cm =1.06 m. v=f... 1
- A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero err… Answer: 4. 0.529 cm Observed reading =0.5+25×0.001=0.525 cm. Zero error is -0.004 cm, so the correction is +0.004 cm. d=0.525-(-0.004)=0.529... 1