** ERRORS**

The result of every measurement by any measuring instrument contains some uncertainty this uncertainty is called errors. The errors in a measurement is equal to the difference between the true value and the measured value of any quantity.

Errors = True value – Measured value

**Absolute Errors:- **the several** **values obtained in an experiment measured are a_{1}, a_{2}, a_{3}….an the arthmetic mean of these values will be

Now error in first measurement = Da_{1} = a_{mean} –a_{1}

Similarly in second measurement = Da_{2 } = a_{mean} –a_{2} and so on

Relative error:Relative error=âˆ†ameanamean

**Combination of errors **(maximum possible error) **Addition or subtraction**

Product or multiplication

Let

**Division **

**(4) Powers**

Let C = a^{n}b^{m}

**CONCEPT WITH EXAMPLES**

**Role of constants **

The radius of sphere is measured to be (2.1 ± 0.5) cm and is surface area with error limits.

**Solution: **

Surface area = A = 4pr^{2}

Now% error

= 47.62 %

**Other way **

**= **4 p r^{2}

**= **55.4 cm^{2}

= 26.4 cm^{2}

Now (A ± DA)

= (55.4±26.4)

**FINDING QUANTITY **

Percentage error in determining of acceleration of gravity with the help of simple Pendulum time period *.* Given error in length of pendulum 4% while in The time period it is 2% then percentage error in g = ?

**Solution**

**LEAST COUNT AND DIFFERENT UNITS**

If a particle of mass m = 25.0 Kg is moving in a circular path of radius r = 50.00 cm with constant speed of v = 10.0 Km/hr them find error in force required by particle.

Now Given

m = 25.0 Kg

= L.c. = 0.1

V = 10.0 Km/h

= L.c = 0.1

DV = 0.1, V = 10

R = 50.00 cm

Dr = 0.001, r = 50cm

**NOTE:** never change units in error.

**Least count **

25.00 – L.c = 0.01

16.50 – 50 –L.c = 0.01

76.03 – 03 –L.c = 0.01

15.0002 –L.c = 0.0001

17.030 – L.c = 0.01

**Solution:**

= 1.21 x 2= 2.42% **Ans.**

**UNIQUE CONCEPT**

Calculate focal length of a spherical Mirror from the following observations Object distance u = (50.1 ± 0.5) cm. and image distance v = (20.1 ± 0.2) cm.

= 14.3 cm

Now

± 0.4 cm

**Note: ** Similar case in resistance connected in parallel combinations

**SIGNIFICANT FIGURES **

The number of significant figures of a numerical quantity is the number of reliably known digits it contains:

**Rules****:**

1. Zeros at the beginning of a number are not significant.

Ex. 0.0523 – there S.f. (5, 2, 3)

**2. **Zeros within a number are significant.

Ex: 2056 –Four S.f. (2, 0, 5, 6)

**3. **Zeros at the end of a number after decimal points are significant.

Ex. 3702.0 –five S.f. (3, 7, 0, 2, 0)