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(1) 20 N

(2) 50 N

(3) 100 N

(4) 2 N

**[AIEEE – 2003] **

**Sol.**(4)

(1) 0.02 (2) 0.03 (3) 0.06 (4) 0.01

**[AIEEE – 2003]**

**Sol.**(3)

(1) 2.0 (2) 4.0 (3) 1.6 (4) 2.5

**[AIEEE – 2004]**

**Sol.**(1)

$(1) \mu_{\mathrm{k}}=1-\frac{1}{\mathrm{n}^{2}}$

$(2) \mu_{\mathrm{k}}=\sqrt{1-\frac{1}{\mathrm{n}^{2}}}$

(3) $\mu_{\mathrm{s}}=1-\frac{1}{\mathrm{n}^{2}}$

(4) $\mu_{\mathrm{s}}=\sqrt{1-\frac{1}{\mathrm{n}^{2}}}$

** [AIEEE – 2005]**

**Sol.**(1)

(1) $2 \sin \phi$

(2) $2 \cos \phi$

(3) 2 tan $\phi$

(4) $\tan \phi$

**[AIEEE – 2005]**

**Sol.**(3)

(1) 800 m (2) 1000 m (3) 100 m (4) 400 m

**[AIEEE – 2005]**

**Sol.**(2)

(1) 4 (2) 1 (3) 2 (4) 3

**[AIEEE – 2011]**

**Sol.**(4)

$=\frac{2 \mu+\mu}{2 \mu-\mu}=\frac{3 \mu}{\mu}=3$

(1) $\frac{1}{3} \mathrm{m}$

(2) $\frac{1}{2} \mathrm{m}$

(3) $\frac{1}{6} \mathrm{m}$

(4) $\frac{2}{3} \mathrm{m}$

**[jee-main-2014]**

**Sol.**(3)

For equilibrium under limiting friction

$\operatorname{mg} \sin \theta=\mu \operatorname{mg} \cos \theta$

$\Rightarrow \tan \theta=\mu$

From the equation of surface $y=\frac{x^{3}}{6}$

slope $=\frac{d y}{d x}=\frac{3 x^{2}}{6}=\tan \theta$

$\Rightarrow \frac{\mathrm{x}^{2}}{2}=\mu=0.5 \Rightarrow \mathrm{x}=1$

So $\mathrm{y}=\frac{1}{6}$

(1) 120 N (2) 150 N (3) 100 N (4) 80 N

**[jee-main-2015]**

**Sol.**(1)

for equllibrrium of A

$\mathrm{f}_{1}=20$

for equllibrrium of B

$\mathrm{f}_{2}=\mathrm{f}_{1}+100$

$\mathrm{f}_{2}=120 \mathrm{N}$

(1) 27.3 kg (2) 43.3 kg (3) 10.3 kg (4) 18.3 kg

** [JEE-(Mains) – 2018]**

**Sol.**(1)

hi

thank you

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