Units and Measurement
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Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
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Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
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The least count (LC) of the screw gauge is the pitch divided by the number of divisions on the circular scale.
LC = Pitch / N = 0.1 cm / 50 = 0.002 cm.
A measurement taken with this instrument must be a multiple of its least count. We check the options:
(A) 2.121 / 0.002 = 1060.5 (Not a multiple)
(B)
The displacement time graph of a particle executing SHM is given in figure: (sketch is schematic and not to scale)
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The displacement time graph of a particle executing SHM is given in figure: (sketch is schematic and not to scale)
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F = MA = m (-ω²x). F=0 at x=0 (t=T/4, 3T/4.). (a) is correct.
Acceleration is max at extremes (t=0, T/2, T.). (b) is correct.
Speed is max at mean position (t=T/4, 3T/4.). (c) is correct.
PE=KE at x=A/√2. t=T/8. (d) is incorrect.
An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
<!-- [if !supportLineBreakNewLine]--><!--[endif]-->(a) the mean free path of the molecules decreases
(b) the mean collision time between the molecules decreases.
(c) the mean free path remains unchanged.
(d) the mean collision time remains unchanged.
An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
<!-- [if !supportLineBreakNewLine]--><!--[endif]-->(a) the mean free path of the molecules decreases
(b) the mean collision time between the molecules decreases.
(c) the mean free path remains unchanged.
(d) the mean collision time remains unchanged.
On increasing the temperature, random velocity of molecules increases, therefore mean collision time between the molecules decreases. But the mean free path remains constant as it is product of velocity and time.? (b) and (c) are correct.
If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be
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If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be
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Y ∝ F? V? A? Y = (E/A)/L²
ML? ¹T? ² ∝ (M¹L¹T? ²)? (L¹T? ¹)? (L²)?
a=1, a+b+2c=-1, -2a-b=-2
a=1, b=0, c=-1
Y = F¹V? A? ¹
The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of cylinder the zero of the vernier scale is between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division)
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The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of cylinder the zero of the vernier scale is between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division)
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Zero error = 0 + 7 × 0.1 = 0.070
Vernier reading = (3.1 + 4 × 0.01) – 0.07 = 3.07
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is (x/100)%. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is (x/100)%. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is
ρ = m/ (4/3πr³). Δρ/ρ = Δm/m + 3Δd/d.
= 6+3×1.5=10.5%. (1050/100)%
