Errors in Measurements

PARAGRAPH FOR QUESTION NO. 8 and 9 If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are $\Delta x$ , $\Delta y$ and $\Delta z$ , respectively, then $z \pm \Delta z = \frac {x \pm \Delta x}{y \pm \Delta y} = \frac {x}{y} \left(1 \pm \frac {\Delta x}{x}\right) \left(1 \pm \frac {\Delta y}{y}\right) ^ {- 1} $ The series expansion for $\left(1\pm\frac{\Delta y}{y}\right)^{-1}$ , to first power in $\Delta y/y$ , is $1\mp(\Delta y/y)$ . The relative errors in independent variables are always added. So the error in z will be $ \varDelta z = z \left(\frac {\varDelta x}{x} + \frac {\varDelta y}{y}\right) $ The above derivation makes the assumption that $\frac{\Delta x}{x} \ll 1$ , $\frac{\Delta y}{y} \ll 1$ . Therefore, the higher powers of these quantities are neglected. In an experiment the initial number of radioactive nuclei is 3000. It is found that $1000 \pm 40$ nuclei decayed in the first 1.0 s. For $|x| << 1$ , $\ln (1 + x) = x$ up to first power in $x$ . The error $\Delta \lambda$ , in the determination of the decay constant $\lambda$ , in $s^{-1}$ , is :-

A steel wire of diameter 0.5 mm and Young's modulus $2 \times 10^{11} N m^{-2}$ carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is..... (Take $g = 10 ms^{-2}$ and $\pi = 3.2$ ).

The following observations were taken for determining surface tension T of water by capillary method: Diameter of capillary, $D = 1.25 \times 10^{-2} m$ Rise of water, $h = 1.45 \times 10^{-2} m$ Using $g = 9.80 \, m/s^{2}$ and the simplified relation $T = \frac{rhg}{2} \times 10^{3} \, N/m$ , the possible error in surface tension is closest to:

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\delta T = 0.01$ second and he measures the depth of the well to be $L = 20$ meters. Take the acceleration due to gravity $g = 10\mathrm{ms}^{-2}$ and the velocity of sound is $300\mathrm{ms}^{-1}$ . Then the fractional error in the measurement, $\delta L / L$ , is closest to

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the $45^{th}$ division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the $25^{th}$ division coincides with the main scale line?

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

If $a, b, c, d$ are inputs to a gate and $x$ is its output, then as per the following time graph, the gate is

Identify the semiconductor devices whose characteristics are given below, in the order $(a)$ , $(b)$ , $(c)$ , $(d)$ :
(a)

(b)

(c)

(d)

There are two vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers ( $C_{1}$ ) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper ( $C_{2}$ ) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers $C_{1}$ and $C_{2}$ respectively, are

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is $T = 2\pi \sqrt{\frac{7(R-r)}{5g}}$ . The values of R and r are measured to be $(60 \pm 1)$ mm and $(10 \pm 1)$ mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

The energy of a system as a function of time t is given as $E(t) = A^{2} \exp(-\alpha t)$ , where $\alpha = 0.2s^{-1}$ . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E(t)$ at t = 5 s is.

The current voltage relation of diode is given by $I = (e^{1000V/T} - 1) \, mA$ , where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring $\pm 0.01V$ while measuring the current of 5 mA at 300 K, what will be error in the value of current in mA?

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

The forward biased diode connection is:

During Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{th}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{th}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is $8 \times 10^{-7} m^{2}$ . The least count of the Vernier scale is $1.0 \times 10^{-5} m$ . The maximum percentage error in the Young's modulus of the wire is

Using the expression $2d \sin \theta = \lambda$ , one calculates the values of d by measuring the corresponding angles $\theta$ in the range 0 to $90^{\circ}$ . The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$ . As $\theta$ increases from $0^{\circ}$ :-

Round off the following number within three significant figures $-4.735 \times 10^{-6} kg$

$4.338 + 4.835 \times 3.88 \div 3.0$ is equal to:

The length, breadth and thickness of a block are given by $l = 12 \text{ cm}$ , $b = 6 \text{ cm}$ and $t = 2.45 \text{ cm}$ . The volume of the block according to the idea of significant figures should be

The vernier of a circular scale is divided in to 30 divisions, which coincides with 29 main scale divisions. If each main scale division is $(1/2)^{\circ}$ , the least count of the instrument is

The pitch of a screw gauge is 0.5 mm and there are 100 divisions on it circular scale. The instrument reads +2 circular scale divisions when nothing is put in-between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and $83^{rd}$ circular scale division coincides with the reference line. Then the diameter of the wire is

The circular divisions of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

An experiment measures quantities x, y, z and then t is calculated from the data as $t = \frac{xy^{2}}{z^{3}}$ . If percentage errors in x, y and z are respectively 1%, 3%, 2%, then percentage error in t is: