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
2.0
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.
The formula provided for the period ($T$) of a simple pendulum is:
$\text{T} = 2 \pi \sqrt{\frac{\text{l}}{\text{g}}}$
Where:
We are given the following values:
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
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 $2.0$.
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.
In an examination 10,000 students appeared. The result revealed the number of students who have:
passed in all five subjects = 5583
passed in three subjects only = 1400
passed in two subjects only = 1200
passed in one subject only = 735
failed in English only = 75
failed in Physics only = 145
failed in Chemistry only = 140
failed in Mathematics only = 200
failed in Bio-science only = 157
The number of students passed in at least four subjects is:
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 ?
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