Get Answers to all your Questions

header-bg qa
Filter By

All Questions

Locus of the midpoint of any focal chord of \mathrm{y}^2=4 \mathrm{ax} is
 

Option: 1

\mathrm{y}^2=\mathrm{a}(\mathrm{x}-2 \mathrm{a})
 


Option: 2

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-2 \mathrm{a})

 


Option: 3

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-\mathrm{a})
 


Option: 4

\text{None of these}


Let the midpoint be \mathrm{P}(\mathrm{h}, \mathrm{k}). Equation of this chord is \mathrm{\mathrm{T}=\mathrm{S}_1}. i.e., \mathrm{\mathrm{yk}-2 \mathrm{a}(\mathrm{x}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. It must pass through (\mathrm{a}, 0)

\mathrm{\Rightarrow 2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. Thus required locus is \mathrm{\mathrm{y}^2=2 \mathrm{ax}-2 \mathrm{a}^2.}

Hence option 2 is correct.

 

View Full Answer(1)
Posted by

SANGALDEEP SINGH

What is the wavelength of a photon with energy 3.0 \times 10^{-19} \text{ J}, according to Planck's quantum theory?

Option: 1

1.15 \times 10^{-7} \text{ m}


Option: 2

2.50 \times 10^{-7} \text{ m}


Option: 3

5.56 \times 10^{-7} \text{ m}


Option: 4

9.23 \times 10^{-7} \text{ m}


C is correct answer,

 

View Full Answer(2)
Posted by

Mohit singh

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


3/4gm/s2

 

View Full Answer(3)
Posted by

Guru G

Calculate the acceleration of block m_1 of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg

 

Option: 1

0.33m/s2


Option: 2

0.66m/s2


Option: 3

1m/s2


Option: 4

1.32m/s2


0.66m/s2

View Full Answer(5)
Posted by

Guru G

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Mercury in water changes to soluble

Option: 1

Dimethyl mercury


Option: 2

diethyl mercury


Option: 3

Hg_2Cl_2


Option: 4

None of these


2

View Full Answer(2)
Posted by

Karumudi.santhoshreddy

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks


In the following the most stable conformation of n-butane is:

Option: 1


Option: 2


Option: 3


Option: 4


4

View Full Answer(2)
Posted by

Jayasri srikanth

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both GM and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

View Full Answer(1)
Posted by

Ajit Kumar Dubey

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

A 100 \; m long wire having cross-sectional area 6.25 \times 10^{-4}m^{2} and Young's modulus is 10^{10}Nm^{-2}  subjected to a load of 250\; N, then the elongation in the wire will be:
Option: 1 4 \times 10^{-3} \mathrm{~m}
Option: 2 6.25 \times 10^{-3} \mathrm{~m}
Option: 3 6.25 \times 10^{-6} \mathrm{~m}
Option: 4 4 \times 10^{-4} \mathrm{~m}

\begin{aligned} & \text { Stress }=\mathrm{y} \text { strain } \Rightarrow \frac{W}{\mathrm{~A}}=\mathrm{y} \frac{\Delta \ell}{\ell} \\ & \Delta \ell=\frac{\mathrm{W} \ell}{\mathrm{yA}} \Rightarrow \Delta \ell=\frac{250 \times 100}{10^{10} \times 6.25 \times 10^{-4}} \\ & \Delta \ell=4 \times 10^{-3} \mathrm{~m} \end{aligned}

Hence, the correct answer is option 1

View Full Answer(1)
Posted by

rishi.raj

A circular loop of radius r is carrying current I\; A. The ratio of the magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is:
Option: 1 2 \sqrt{2}: 1
Option: 2 1: 3 \sqrt{2}
Option: 3 1: \sqrt{2}
Option: 4 3 \sqrt{2}: 2

Magnetic field at centre of coil B_1=\frac{\mu_0 I}{2 r}

on the axis at

 x=r \Rightarrow B_2=\frac{\mu_0 \mathrm{Ir}^2}{2\left(r^2+x^2\right)^{3 / 2}}

\begin{aligned} & \mathrm{B}_2=\frac{\mu_0 \mathrm{Ir}^2}{2\left(\mathrm{r}^2+\mathrm{r}^2\right)^{3 / 2}} \\ & \mathrm{~B}_2=\frac{\mu_0 \mathrm{I}}{2(2 \sqrt{2} r)} \\ & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=2 \sqrt{2} \end{aligned}

\(2 \sqrt{2}: 1\)

 

View Full Answer(1)
Posted by

rishi.raj

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

filter_img