Fundamentals of Physics Test - 2

The least count of an instrument is the smallest measurement that can be accurately taken from that particular instrument.

A Travelling microscope is a compound microscope that is fitted on a vertical scale. It carries a vernier scale along the main scale and can be moved upward or downward.

So, the smallest division on the main scale of the travelling microscope is = 0.5 mm

Number of divisions on vernier scale = 50 divisions

The least count of a travelling microscope can be calculated as

Least count (LC) = Smallest division on main scale/Number of divisions on the vernier scale

LC = 0.5/50

LC = 0.01 mm

LC = 0.001 cm

Hence, the correct option is (A).

Distance \(=160\) km

Time taken \(=4\) hr

Speed \(=?\)

Speed = Distance covered/Time taken \(=160 / 4 =40\) km/hr

Hence, the correct option is (A).

A Vernier scale on a caliper may have a least count of 0.1 mm while a micrometer may have a least count of 0.01 mm. The least count error occurs with both systematic and random errors.

A simple harmonic oscillator, consisting of a weight attached to one end of a spring, is shown. The other end of the spring is connected to a rigid support such as a wall. If the system is left at rest at the equilibrium position then there is no net force acting on the mass. However, if the mass is displaced from the equilibrium position, the spring exerts a restoring elastic force that obeys Hooke's law.

The zero error is defined as such a condition when a measuring instrument registers a reading when there should not be any reading. In the case of vernier callipers, it occurs when a zero on the main scale does not coincide with a zero on vernier scale.

So vernier callipers are free from zero error if zero on the main scale coincide with zero on vernier scale.

Least count \(, L C=1 M S D-1 V S D\)

Here \(10 V S D=9 M S D\) or \(1 V S D=(9 / 10) M S D\)

Thus, \(L C=(1-9 / 10) 1 M S D=1 / 10 \times 0.1 \mathrm{cm}=0.01 \mathrm{cm}\) (as \(1 \mathrm{MSD}=0.1 \mathrm{cm}\) from diagram)

From diagram, main scale reading, \(M S R=4.3 \mathrm{cm}\) and vernier scale reading, \(V S R=7\) (because division \(7 \text { coincide with any line on main scale })\)

The diameter of the metal ball \(=M S R+(V S R \times L C)=4.3+(7 \times 0.01)=4.37 \mathrm{cm}\)

Radius of the ball \(=4.37 / 2=2.185 \mathrm{cm}=2.18 \mathrm{cm}\)

The timekeeping elements in all clocks, which include pendulums, balance wheels, the quartz crystals used in quartz watches, and even the vibrating atoms in atomic clocks, are in physics called harmonic oscillators. The reason harmonic oscillators are used in clocks is that they vibrate or oscillate at a specific resonant frequency or period and resist oscillating at other rates. However, the resonant frequency is not infinitely 'sharp'. Around the resonant frequency there is a narrow natural band of frequencies (or periods), called the resonance width or bandwidth, where the harmonic oscillator will oscillate. In a clock, the actual frequency of the pendulum may vary randomly within this resonance width in response to disturbances, but at frequencies outside this band, the clock will not function at all.

Given: \(50 \mathrm{VSD} \equiv 49 \mathrm{MSD}\)

Hence, \(1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}\)

Least count of the instrument \(=M S D-V S D=2\left(1-\frac{49}{50}\right)\)

\((\because M S D \text { is given as } 2 m m)\)

Hence, least count of vernier calipers \(=2\left(\frac{1}{50}\right)=\frac{1}{25} \mathrm{mm}\)

Least count, \(L C=0.01 \mathrm{cm}\)

Here the negative zero error of callipers \(=5 \times 0.01=0.05 \mathrm{cm}\) and it should be added to final reading.

Main scale reading, \(M S R=2.4 \mathrm{cm}\) and \(V S D=6\)

So total reading means the diameter of sphere \(R=M S R+(V S D \times L C)+\) zero error \(=2.4+\) \((6 \times 0.01)+0.05=2.51 \mathrm{cm}\)

Pitch of the screw gauge P = 0.5 mm

Number of divisions on circular scale N = 200

So, least count L.C. = $\frac{P}{N}=\frac{0.5mm}{200}=0.0025=0.00025 cm$

Time period of oscillation of a simple pendulum is given by: 

A seconds pendulum is a pendulum whose period is precisely two seconds; one second for a swing in one direction and one second for the return swing. i.e.

T = 2s

and T = $2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$

⇒ $l=\frac{T^{2}g}{4{\mathrm{\pi }}^2}=\frac{2^2\times 980}{4{\mathrm{\pi }}^2}=99.29cm\cong 99.2cm$

The metric scale is the system of measurement used in the metric system. In the metric scale, 10 centimeters are equal to one decimeter, 10 decimeters are equal to one meter and 1,000 meters are equal to one kilometer.

(a) Least count of vernier callipers

\(=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{MSD}-\frac{19}{20} \mathrm{MSD}\)

\(=\frac{1}{20} M S D=\frac{1}{20} m m=\frac{1}{200} \mathrm{cm}=0.005  \mathrm{cm}(\because 1 \mathrm{MSD}=1 \mathrm{mm})\)

(b) Least count of screw gauge 

\(=\frac{P i t c h}{N o . o f \text { divisions on circular scale}}\)

\(\frac{1}{100} m m=\frac{1}{100} \mathrm{cm}=0.001 \mathrm{cm}\)

(c) Least count of spherometer 

\(=\frac{\text {Pitch}}{\text {No.of divisions on circular scale}}\)

\(\frac{0.1 \mathrm{mm}}{100}=\frac{1}{1000} \mathrm{mm}=0.0001 \mathrm{cm}\)

LC. of the screw gauge \(=\frac{P i t c h}{\text {Number of divisions on circular scale}}=\frac{1}{100}=0.01 \mathrm{mm}\)

Given : Main scale reading \(=1 \mathrm{mm}\)

Circular scale reading \(=0.01 \times 47=0.47\)

\(\therefore\) Diameter of wire \(=\) Total reading \(=1.47 \mathrm{mm}=0.147 \mathrm{cm}\)

Curved surface area \(=\pi \times d l=\pi \times(0.147)(5.6)=2.586 \approx 2.6 \mathrm{cm}^{2}\)

The upper jaws are designed with parallel outer surfaces for use in measuring inside diameters. The lower jaws have parallel inner surfaces, and are used for outside diameter measurements.

Number of division on main scale n = 10

Number of division on vernier scale m = 20

Main scale division 1 M.S.D. = 1 mm

Least count of vernier caliper L.C. = M.S.D. $\times \frac{n}{m}=1mm\times \frac{10}{20}=0.5mm$

Main scale reading M.S.R. = 35 mm

Coinciding division n₁ = 5

Diameter of wire d = M.S.R. + n₁ × L.C. = 35 + 0.5 × 0.5 = 35.25 mm

Conductivity has relation, σ = \(\frac{l}{A R}=\frac{l I}{A V}\).

V has dimension [ML²T^-3I^-1]. Hence dimension of σ is [M^-1L^-3T³I²].

From the figure, 0 of vernier callipers has gone past \(3+0.3=3.3 \mathrm{cm}\) of main scale. Also, \(6^{\mathrm{th}}\) division line of vernier scale matches with line of main scale.

So, total reading is \(3.3+6 \times 0.01=3.36 \mathrm{cm}\)

The statement, "Each volumes of two different substances have equal masses" is false. They have different masses.

Hence, the correct option is B.