Question

An electro accelerates in a straight line at 2×10ppwer 10 m/s²in a cathode ray tube. If it's initial speed is 4×10⁴ m/s. Calculate it's speed after 3×10 power minus 5 seconds.

Answer 1

64 × 10⁴ m/s is the required answer.

GIVEN:

• Acceleration of electron, a = 2 × 10 ¹⁰ m/s²
• Initial speed of electron , u = 4 × 10⁴ m/s
• Time, t = 3 × 10^-5 sec

TO FIND:

• Speed after the given time ( final velocity) , v

FORMULA:

According to first equation of motion, v = u + at

SOLUTION:

$v = u + at \\ \\ = (4 \times {10}^{4}) + (2 \times {10}^{10} \times 3 \times {10}^{ - 5} ) \\ \\ = (4 \times {10}^{4} ) + (6 \times {10}^{5} ) \\ \\ = (4 \times {10}^{4}) + (60 \times {10}^{4} ) \\ \\ = {10}^{4} (4 + 60) \\ \\ v= 64 \times {10}^{4} \: \frac{m}{s}$

ANSWER:

The speed of electron after the given time is 64 × 10⁴ m/s

OTHER FORMULAE:

• First equation of motion, v = u + at

• Second equation of motion , s = ut + (at²)/2

• Third equation of motion , v²- u² = 2as

• These equations of motions can be applied only if acceleration is constant.

CATHODE RAY TUBE:

• It is used to generate electrons and strike the screen with high velocity.

• The deflection of electrons in the cathode ray rube are modified using electric and magnetic field vectors, which are perpendicular to each other.