CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will :



A
increase by 19%
loader
B
decrease by 19%
loader
C
increase by 11%
loader
D
decrease by 11%
loader

Solution

The correct option is D increase by 11%
The time period of oscillation is given by
$$T = 2 \pi \sqrt {\dfrac{I}{MB}} $$
$$ m_2 = m_1 - 0.19 m_1 $$
$$ m_2 = 0.81 m_1 $$

$$ \dfrac{T_2}{T_1} = \sqrt{ \dfrac{m_1}{m_2} } $$

$$ \dfrac{T_2}{T_1} = \dfrac{1}{0.9} $$

$$ \dfrac{ \Delta T}{T_1} \times 100 \approx 11 \%$$

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image