All Exams Test series for 1 year @ ₹349 only
Question

The period of a pendulum is given as T = 2 π (l/g)1/2 where g = 9.81 m/s2 and π = 3.1416. The period of a pendulum of length 1 m correct to the first place of decimal in seconds is

The correct answer is

2.0

Pendulum Period Calculation

The question asks us to find the period of a simple pendulum with a given length, using a standard formula and specific values for gravity and pi. We need to calculate the period and round the result to the first decimal place.

Pendulum Formula

The formula provided for the period ($T$) of a simple pendulum is:

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

Where:

  • $\text{T}$ is the period of the pendulum (in seconds)
  • $\pi$ is a mathematical constant
  • $\text{l}$ is the length of the pendulum (in meters)
  • $\text{g}$ is the acceleration due to gravity (in m/s²)

Given Values

We are given the following values:

  • Length of the pendulum, $\text{l} = 1 \text{ m}$
  • Acceleration due to gravity, $\text{g} = 9.81 \text{ m/s}^2$
  • Value of pi, $\pi = 3.1416$

Calculating the Pendulum Period

Now, we substitute the given values into the formula for the period:

$\text{T} = 2 \times 3.1416 \times \sqrt{\frac{1}{9.81}}$

First, calculate the value inside the square root:

$\frac{1}{9.81} \approx 0.1019368$

Next, calculate the square root of this value:

$\sqrt{0.1019368} \approx 0.319275$

Now, multiply the terms together:

$\text{T} = 2 \times 3.1416 \times 0.319275$

$\text{T} = 6.2832 \times 0.319275$

$\text{T} \approx 2.00581$ seconds

Rounding the Period

The question asks for the period correct to the first place of decimal. Our calculated value is approximately $2.00581$.

To round to the first decimal place, we look at the second decimal digit. If it is 5 or greater, we round up the first decimal digit. If it is less than 5, we keep the first decimal digit as it is.

In $2.00581$, the first decimal digit is 0, and the second decimal digit is 0. Since 0 is less than 5, we keep the first decimal digit as 0.

So, $2.00581$ rounded to the first decimal place is $2.0$.

Final Answer

The period of the pendulum of length 1 m, correct to the first place of decimal, is 2.0 seconds.

Parameter Value
Pendulum Length (l) 1 m
Gravity (g) 9.81 m/s²
Pi ($\pi$) 3.1416
Calculated Period (T) $\approx 2.00581$ s
Rounded Period (T) 2.0 s

The calculated period of the pendulum, rounded to one decimal place, matches one of the provided options.

Was this answer helpful?

Important Questions from Numerical Ability

  1. Four identical cones with base diameter of 10 cm are compactly placed inside a box in upright position. What will be the area of square (in cm2) formed by connecting tips of the cones?
  2. How many hollow spheres having inner radius of 1 cm can be completely filled by transferring water from a completely filled hollow sphere having inner diameter of 20 cm ?

  3. The sides a, b and c of a Δ ABC satisfy the equation (a – 8)2 + (b - 15)2 + (c - 17)2 = 0. Then Δ ABC is

  4. In the given subtraction problem, each letter represents a digit between 0 and 9.

    TAS5
    -RSR
    2TA9

    The values of R, A and T are, respectively
  5. A milk vendor has 50 L of milk and supplies 5 L to every customer. After each transaction he adds 5L of water. What is the percentage of milk contained in a litre of solution purchased by the fifth customer?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App