Question

A saturated specimen of clay was immersed in mercury and displaced volume of 30 cc and weight of sample obtained was 45 gm. After oven drying for 48 hours, weight reduced to 30 gm. While the volume came down to 18cc. The specific gravity of clay specimen is

• A. 2.5
• B. 3.2
• C. 2
• D. 1.75

Answer 1

The correct option is C 2

$$\begin{array}{cc}Given:W_{sat}=45gm& W_s=W_d=30gm\\ V_{sat}=30cc& V_d=18cc\end{array}$$

$\therefore W_s=\frac{(W_{sat}-W_s)-(V_{sat}-V_d).{\gamma }_w}{W_s}\times 100$

$=\frac{(45-30)-(30-18)\times 1}{30}\times 100$

$\therefore W_s=10\%$

$\therefore W_s=[\frac{1}{(G_m{)}_{dry}}-\frac{1}{G}]\times 100$

$(G_m{)}_{dry}=\frac{{\gamma }_d}{{\gamma }_w}$

${\gamma }_d=\frac{W_d}{V_d}=\frac{30}{18}$

${\gamma }_d=1.67g/cc$

$(G_m{)}_{dry}=\frac{1.67}{1}=1.67$

$10\%=[\frac{1}{1.67}-\frac{1}{G}]\times 100$

$\therefore G=2.00$