The length of two rods are recorded as `l_(1) = (25.2 pm 0.1)` cm and `l_(2) = (16.8 pm 0.1)` cm . Find their combined length .

The lengths of two rods are recorded as l_(1)=25.2pm0.1cm and l_(2)=16.8pm0.1cm . Find the sum of the lengths of the two rods with the limits of error.

If l_(1) = (10.0 pm 0.1) cm and l_(2) = (9.0 pm 0.1) cm , find the their sum , difference and error in each .

If l_(1) = (10.0 pm 0.1) cm and l_(2) = (9.0 pm 0.1) cm , find the their sum , difference and error in each .

The lengths l_(1) and l_(2) of two rods are recorded as (25.2+0.1) cm and (16.8+0.1) cm. To find the sum of the lengths of the rods.

The lengths l_(1) and l_(2) of two rods are recorded as (25.2+0.1) cm and (16.8+0.1) cm. To find the sum of the lengths of the rods.

Two rods have lengths measured as (1.8 pm 0.2) m and (2.3 pm 0.1) m. Calculate their combined length with error limits.

Two rods have lengths measured as (1.8 pm 0.2) m and (2.3 pm 0.1) m. Calculate their combined length with error limits.

Two rods have lengths measured as (1.8 pm 0.2) m and (2.3 pm 0.1) m. Calculate their combined length with error limits.

The lengths of the two pieces of wire are l_1 =(35.2 +- 0.1) cm and l_2 = (47.4 +- 0.2) cm. what is the total length of wire with error limits ?

The diameter of two cylingers are found to be d_(1)= (2 pm 0.01)cm" and "d_(2)=(3.1 pm 0.03)cm . Find the difference between both the diameters along with the error limits.