MCQ
in radio theraphy, $X-$ rays are used to
- ADetect bone fractures
- ✓Treat cancer by controlled exposure
- CDetect heart diseases
- DDetect fault in radio receiving circuits
✓
Answer
Correct option: B.
Treat cancer by controlled exposure
b
(b) Refer to the application of $X-$ rays.
(b) Refer to the application of $X-$ rays.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A diode detector is used to detect an amplitudemoduiated wave of $60\%$ modulation by using a condenser of capacity $250$ picofarad in parallel with a load resistance $100$ kilo ohm. Find the maximum modulated frequency which could be detected by it.
→For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
→If a particle of charge ${10^{ - 12}}\,coulomb$ moving along the $\hat x - $ direction with a velocity ${10^5}\,m/s$ experiences a force of ${10^{ - 10}}\,newton$ in $\hat y - $ direction due to magnetic field, then the minimum magnetic field is
→Who first proposed the light as an electromagnetic wave?
When the length of a microscope tube increases, its magnifying power
The angular resolution of a $10 \;cm$ diameter telescope at a wavelength of $5000 \;\mathring A$ is of the order
→A frog can be deviated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as
What is the atomic mass $(u)$ of calcium?
→Assertion : Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it.
Reason : The average velocity of free electron is zero.
The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} Wb / m ^{2}$. If the angle of dip at that place is $60^{\circ}$, then the value of vertical component of the earth's magnetic field will be ........ $\times 10^{-4}\;W b / m^{2}$
→