Home
Class 11
PHYSICS
A student measures the time period of 10...

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

A

`92 pm 2`s

B

`92 pm 5.0`s

C

`92 pm 1.8`s

D

`92 pm3`s

Text Solution

Verified by Experts

The correct Answer is:
A
Average value of time period,
`barx=(undersetoverset sum(i=1)(4)x_(i))/(N)=(90+91+95+92)/(4)=1.5`
Average maximum error
`in=(undersetoversetsum(i=1)(4 )|barx-x_(i)|)/(N)=(2+1+3+0)/(4)=1.5`
As the clock is able to measure minimum time of 1 s, the reported mean time = `92 pm 2` s.
Promotional Banner

Topper's Solved these Questions

  • MEASUREMENT AND DIMENSION OF PHYSICAL QUANTITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (AIPMT)|1 Videos

  • MEASUREMENT AND DIMENSION OF PHYSICAL QUANTITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (NEET)|2 Videos

  • MEASUREMENT AND DIMENSION OF PHYSICAL QUANTITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (WBJEE )|4 Videos

  • KINETIC THEORY OF GASES

    CHHAYA PUBLICATION|Exercise CBSE Scanner|9 Videos

  • NATURE OF VIBRATION

    CHHAYA PUBLICATION|Exercise EXERCISE (CBSE SCANNER)|2 Videos

Similar Questions

Explore conceptually related problems

What is the time period of a simple pendulum of effective length 60 cm? g=980cm*s^(-2) .

The time period of a simple pendulum is 2 s. The mass of its bob is 10 g and the bob is oscillating with an amplitude of 4 cm. What is the total mechanical energy of the bob?

Time period of a simple pendulum is 2 s. If its length is doubled, then the new time period will be

The period of oscillation of a simple pendulum is T = 2pisqrt((L)/(g)). Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is