JEE Mains · Physics · STD 12 - 3. current electricity
When two resistance \(R_1\) and \(R_2\) connected in series and introduced into the left gap of a meter bridge and a resistance of \(10 \Omega\) is introduced into the right gap, a null point is found at \(60 cm\) from left side. When \(R_1\) and \(R_2\) are connected in parallel and introduced into the left gap, a resistance of \(3 \Omega\) is introduced into the right-gap to get null point at 40 cm from left end. The product of \(R_1 R_2\) is \(.............\Omega\)
- A \(31\)
- B \(30\)
- C \(32\)
- D \(33\)
Answer & Solution
Correct Answer
(B) \(30\)
Step-by-step Solution
Detailed explanation
\(\frac{ R _1+ R _2}{10}=\frac{60}{40}=\frac{3}{2} \Rightarrow R _1+ R _2=15\) Now \(\frac{ R _1 R _2}{\left( R _1+ R _2\right) \times 3}=\frac{40}{60}=\frac{2}{3} \Rightarrow R _1 R _2=30\)
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Physics
- A screw gauge has \(50\) divisions on its circular scale. The circular scale is \(4\) units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of \(0.5\, mm\) is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectivelyJEE Mains 2020 Medium
- A particle initially at rest starts moving from reference point. \(\mathrm{x}=0\) along \(\mathrm{x}\)-axis, with velocity \(v\) that varies as \(v=4 \sqrt{\mathrm{x} m} / \mathrm{s}\). The acceleration of the particle is _______ \( \mathrm{ms}^{-2}\).JEE Mains 2024 Hard
- In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ \(\mu \mathrm{C}\).
JEE Mains 2024 Hard - A small bob A of mass m is attached to a massless rigid rod of length 1m pivoted at point P and kept at an angle of \( 60^{\circ} \) with vertical as shown in figure. At distance of 1m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in figure. If bob B just manages to complete the circular path of radius R upto a point Q after being hit elastically by both A, then radius R is ___________m.
JEE Mains 2026 Hard - If the acceleration due to gravity experienced by a point mass at a height \(h\) above the surface of earth is same as that of the acceleration due to gravity at a depth a depth \(\alpha\) h \(\left(h \ll R_{e}\right)\) from the earth surface. The value of \(\alpha\) will be\(....\)(use \(R _{ e }=6400\,km\) )JEE Mains 2022 Medium
- A zener diode having zener voltage \(8\, {V}\) and power dissipation rating of \(0.5\, {W}\) is connected across a potential divide arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance \({R}_{{p}}\) is \(....\,\Omega\)
JEE Mains 2021 Medium
More PYQs from JEE Mains
- In a vernier callipers, each \(cm\) on the main scale is divided into \(20\) equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be \(\dots \; \times 10^{-2} \;mm\)JEE Mains 2022 Medium
- The sum of possible values of \(x\) for \(\tan ^{-1}( x +1)+\cot ^{-1}\left(\frac{1}{ x -1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)\) isJEE Mains 2021 Hard
- The first member of the Balmer series of hydrogen atom has a wavelength of \(6561\; \mathring A\). The wavelength of the second member of the Balmer series (in \(nm\)) isJEE Mains 2020 Medium
- Let the directrix of the parabola \(P: y^2 = 8x\), cut \(x\)-axis at the point \(A\). Let \(B(\alpha, \beta)\), \(\alpha > 1\), be a point on \(P\) such that the slope of \(AB\) is \(3/5\). If \(BC\) is a focal chord of \(P\), then six times the area of \(\triangle ABC\) is :JEE Mains 2026 Medium
- A uniform rod of length \('l'\) is pivoted at one of its ends on a vertical shaft of negligible radius When the shaft rotates at angular speed \(\omega\) the rod makes an angle \(\theta\) with it (see figure). To find \(\theta\) equate the rate of change of angular momentum (direction going into the paper ) \(\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta\) about the centre of mass \((CM)\) to the torque provided by the horizontal and vertical forces \(F_{H}\) and \(F_{V}\) about the CM. The value of \(\theta\) is then such that:
JEE Mains 2020 Hard - The coefficient of \({x^5}\) in the expansion of \({{{x^2} + 1} \over {({x^2} + 4)(x - 2)}}\) isJEE Mains 2023 Hard