JEE Mains · Physics · STD 12 - 3. current electricity
A regular hexagon is formed by six wires each of resistance \(r \Omega\) and the corners are joined to centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be
- A \(\frac{4}{5}r\)
- B \(\frac{5}{8}r\)
- C \(\frac{3}{4}r\)
- D \(\frac{3}{5}r\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{5}r\)
Step-by-step Solution
Detailed explanation
\(R_{e q}=\frac{2 R \times \frac{4 R}{3}}{2 R+\frac{4 R}{3}}=\frac{8 R^2}{10 R}=\frac{4}{5} R\)
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 conducting circular loop is placed in \(X - Y\) plane in presence of magnetic field \(\overrightarrow{ B }=\left(3 t ^{3} \hat{ j }+3 t ^{2} \hat{ k }\right)\) in SI unit. If the radius of the loop is \(1 m\), the induced emf in the loop, at time, \(t =2\,s\) is \(n \pi V\). The value of \(n\) is.JEE Mains 2022 Medium
- A series \(AC\) circuit containing an inductor \((20\,mH),\) a capacitor \((120\,\mu F)\) and a resistor \((60\, \Omega )\) is driven by an \(AC\) source of \(24\,V/50\,Hz.\) The energy dissipated in the circuit in \(60\,s\) isJEE Mains 2019 Medium
- In a common emitter amplifier circuit using an \(n-p-n\) transistor, the phase difference between the input and the output voltages will be.....\(^o\)JEE Mains 2017 Easy
-

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \(R_p=1 \Omega\) as shown in the figure. An external resistance of \(R_e=2 \Omega\) is connected via the sliding contact. The electric current in the circuit is :JEE Mains 2025 Medium - 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
- Match the List-I with List-II
Choose the correct answer from the options given below:List-I List-II A Radio-wave I is produced by Magnetron value B Micro-wave II Due to change in the vibrational modes of atoms C Infrared-wave II Due to inner shell electrons moving from higher energy level to lower energy level D X-ray IV Due to rapid acceleration of electrons JEE Mains 2026 Hard
More PYQs from JEE Mains
- The pressure wave, \(P = 0.01\,sin\,[1000t -3x]\,Nm^{-2},\) corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is \(0\,^oC.\) On some other day when temperature is \(T,\) the speed of sound produced by the same blade and at the same frequency is found to be \(336 \,ms^{-1}\). Approximate value of \(T\) is .... \(^oC\)JEE Mains 2019 Medium
- The Coefficient of \(x ^{-6}\), in the expansion of \(\left(\frac{4 x}{5}+\frac{5}{2 x^2}\right)^9\), is \(........\).JEE Mains 2023 Hard
- The sum of all the integral values of \(p\) such that the equation \(3\sin^2 x + 12\cos x - 3 = p\), \(x \in \mathbb{R}\), has at least one solution, is:JEE Mains 2026 Medium
- Let a circle \(\mathrm{C}\) of radius \(1\) and closer to the origin be such that the lines passing through the point \((3,2)\) and parallel to the coordinate axes touch it. Then the shortest distance of the circle \(\mathrm{C}\) from the point \((5,5)\) is :JEE Mains 2024 Hard
- \(ABC\) is a plane lamina of the shape of an equilateral triagnle. \(D,\) \(E\) are mid points of \(AB\), \(AC\) and \(G\) is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through \(G\) and perpendicular to the plane \(ABC\) is \(I _{0} .\) If part \(ADE\) is removed, the moment of inertia of the remaining part about the same axis is \(\frac{ NI _{0}}{16}\) where \(N\) is an integer. Value of \(N\) is
JEE Mains 2020 Hard - A microscope was initially placed in air (refractive index \(1)\). It is then immersed in oil (refractive index \(2)\). For a light whose wavelength in air is \(\lambda\), calculate the change of microscope's resolving power due to oil and choose the correct option.JEE Mains 2022 Medium