When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises uptocertain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by __ %.

We need to find the percentage change in capillary rise height when the inner radius, density, and surface tension each decrease by 1%.

The height of liquid rise in a capillary tube is given by $$h = \frac{2T\cos\theta}{\rho g r}$$, where $$T$$ is the surface tension, $$\theta$$ is the contact angle, $$\rho$$ is the liquid density, $$g$$ is acceleration due to gravity, and $$r$$ is the inner radius of the tube.

Taking logarithms and differentiating (with $$\theta$$ and $$g$$ held constant) gives $$\ln h = \ln(2\cos\theta) + \ln T - \ln \rho - \ln g - \ln r$$ and hence $$\frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r}.$$

Since each of $$T$$, $$\rho$$, and $$r$$ decreases by 1%, we have $$\frac{\Delta T}{T} = -1\%$$, $$\frac{\Delta \rho}{\rho} = -1\%$$, and $$\frac{\Delta r}{r} = -1\%$$, so that $$\frac{\Delta h}{h} = (-1\%) - (-1\%) - (-1\%) = -1\% + 1\% + 1\% = +1\%.$$

Decreasing the surface tension ($$T$$) by 1% reduces $$h$$ by 1% (since $$h \propto T$$), while decreasing the density ($$\rho$$) or the radius ($$r$$) by 1% increases $$h$$ by 1% in each case (since $$h \propto 1/\rho$$ and $$h \propto 1/r$$). The net effect is $$-1 + 1 + 1 = +1\%$$ increase in the capillary rise height.

The correct answer is Option 3: +1.