It's a Question of your Career!

Hi!

Please specify your state/board, this will help us answer you with the best resources.

Meanwhile, check the link below, apply relevant filters, find suitable resources:

https://school.careers360.com/download/ebooks-and-sample-papers

Hope it helps!

The link to download the Karnataka school exam model paper is given here:

https://school.careers360.com/boards/kseeb/karnataka-sslc-10th-model-papers-2026-27

Given:

Density of sphere, $\rho_s=8 \mathrm{~g} / \mathrm{cm}^3$
First medium: specific gravity $=2$, so $\rho_1=2 \mathrm{~g} / \mathrm{cm}^3$
Second medium: specific gravity $=4$, so $\rho_2=4 \mathrm{~g} / \mathrm{cm}^3$
First angle: $90^{\circ}$
Second angle: $120{ }^{\circ}$
Dielectric constants: $k$ and $k^{\prime}$

For two identical charged spheres,

$
\tan \frac{\theta}{2}=\frac{F_e}{W_{\text {effective }}}

Hello Student,

To find the X-component of the force on -q1, we need to know the positions of -q1, q2 and -q3, along with the distances/angles shown in the figure. The X-component is obtained by calculating the Coulomb force due to each charge and then resolving the forces along the

$
\begin{aligned}
&\text { After an elongation of r, the separation between the charges will be ' } 2 r^{\prime}\\
&\begin{gathered}
\frac{1}{4 \pi \epsilon_0} \times \frac{q^2}{2 r}=\frac{1}{2} k r^2 \\
k=\sqrt{\frac{q^2}{4 \pi \epsilon_0 r^3}}=\frac{q}{2 r} \sqrt{\frac{1}{\pi \epsilon_0 r}}
\end{gathered}
\end{aligned}
$

Hey there,

If you mean how to get/download the e-books for Class 9 from CM SHRI Schools, the exact portal depends on the state because “CM SHRI School” is used for different state programmes.

If you mean Delhi CM SHRI Schools, tell me that and I can give you the

Hi Swetha,

Sorry, we do not cover this exam if you need exam related information you can check the below link.

https://school.careers360.com/schools/asian-christian-high-school-hosur

if you need any other information and study resources please let us know.

Hi Rajwati Km

Let the two numbers be x and y.

Given:

• Difference = 5

• x-y= 5

• Product = 24

• xy=24

From x−y=5:

x=y+5

Substitute in xy=24

y(y+5)=24

y^2+5y−24=0

Factorising:

(y+8)(y−3)=0

So, y=3 or y=-8.

Therefore, the two numbers are 8 and 3 .

UP NEET UG private colleges ka Round 2 cutoff fixed nahi hota. It depends on NEET rank/score, category, college preference, available seats and number of candidates participating in counselling. Round 2 mein vacant seats ke karan cutoff Round 1 se change ho sakta hai. Aap apna NEET score/rank, category aur

Hey there,

If you mean TG EAPCET counselling, then yes, there is a provision to modify options after freezing, but it depends on the phase and the official schedule. The counselling instructions state that frozen options can be changed through the designated process/help-line centre when the modification opportunity is provided.